CP-9 Differential equations — revision question pack

9 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section CP-9. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

CP-9.1 · Find and use an integrating factor to solve differential equations of form dy/dx + P(x)y = Q(x) and recognise when it is appropriate to do so.

Explanation

A first-order linear differential equation must first be written as $y'+P(x)y=Q(x)$ . Its integrating factor is $I(x)=e^{\int P(x)\,dx}$ , which may be quoted.
Multiplying every term by $I$ makes the left side the product derivative $(Iy)'$ , so integration gives $Iy=\int I Q\,dx+C$ and division by $I$ gives the general solution.
This method is appropriate only when $y$ and $y'$ occur linearly and $P,Q$ depend on the independent variable alone.
If the coefficient of $y'$ is not $1$ , divide through before forming $I$ .
Examiners expect the integrating factor, the product-derivative step and the arbitrary constant to be shown.

Worked example

Solve $x\dfrac{dy}{dx}+2y=x^3$ for $x>0$ , given that $y(1)=2$ .

1.Divide by $x$ : $y'+\dfrac2x y=x^2$ , so $I=e^{\int2/x\,dx}=x^2$ .
2.Multiply through: $(x^2y)'=x^4$ .
3.Integrate: $x^2y=\dfrac{x^5}{5}+C$ , hence $y=\dfrac{x^3}{5}+Cx^{-2}$ .
4. $y(1)=2$ gives $C=\dfrac95$ .

Answer: $y=\dfrac{x^3}{5}+\dfrac{9}{5x^2}$ for $x>0$ .

Common mistakes

Don't use $e^{\int2\,dx}$ before dividing the equation by $x$ .
Don't multiply only the $y$ -terms by the integrating factor and leave $Q(x)$ unchanged.
Don't integrate $(Iy)'$ but omit the arbitrary constant.

Exam tip

Write the equation in monic linear form before stating the integrating factor; this secures the method logic.

Tier 1 · Easy

1.
Find an integrating factor for $\dfrac{dy}{dx}+3y=x$ .
(2)
(Total for Question 1 is 2 marks)
2.
Solve $\dfrac{dy}{dx}+2y=e^{-2x}$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
For $x>0$ , solve $\dfrac{dy}{dx}+\dfrac2x y=x^2$ .
(5)
(Total for Question 1 is 5 marks)
2.
A linear equation has integrating factor $e^{x^2}$ and its solution satisfying $y(0)=2$ is $y=e^{-x^2}(x^3+2)$ . Determine the equation in the form $y'+P(x)y=Q(x)$ and verify by substitution that $y$ satisfies your equation.
(4)
(Total for Question 2 is 4 marks)
3.
Find an integrating factor and hence solve $y'+y\cos x=\cos x$ subject to $y(0)=3$ . Hence find $y(\pi/2)$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
On $0\leq x<\pi/2$ , solve $\dfrac{dy}{dx}+2\tan x\,y=\sin x$ subject to $y(0)=2$ . Hence find $y(\pi/3)$ .
(7)
(Total for Question 1 is 7 marks)
2.
Solve $\dfrac{dy}{dx}+\dfrac{2x}{1+x^2}y=\dfrac{x}{(1+x^2)^2}$ subject to $y(1)=1$ . Hence find the exact value of $y$ when $x=\sqrt3$ .
(7)
(Total for Question 2 is 7 marks)
3.
For $x>0$ , solve $x\dfrac{dy}{dx}+(x+1)y=x^2e^{-x}$ subject to $y(1)=4/(3e)$ . Find $y(2)$ and state the limiting value of $y$ as $x\to\infty$ .
(7)
(Total for Question 3 is 7 marks)
4.
For $x>-1$ , solve $y'+\dfrac{1}{x+1}y=x-1$ subject to $y(0)=k$ . Determine the value of $k$ for which the curve touches the $x$ -axis, and for this value of $k$ find the exact value of $x>0$ at which $y$ is least.
(7)
(Total for Question 4 is 7 marks)
5.
Two solutions $y_1$ and $y_2$ of $y'+2y\tanh x=3\tanh x$ satisfy $y_1(0)=4$ and $y_2(0)=1$ . Find both solutions, the area between their graphs for $0\leq x\leq\ln2$ , and their limiting separation as $x\to\infty$ .
(7)
(Total for Question 5 is 7 marks)

CP-9.2 · Find both general and particular solutions to differential equations.

Explanation

A general solution represents a family of solution curves and contains the correct number of independent arbitrary constants: normally one for a first-order equation and two for a second-order equation. A particular solution results when initial or boundary conditions determine all those constants.
Conditions involving derivatives must be applied to the derivative of the complete general solution.
The phrase particular integral has a different meaning: it is one trial response to a non-homogeneous forcing term, not a solution selected by initial data.
The specification also expects sketches of members of a solution family.
Examiners therefore look for constants in the general solution, correct use of every condition, and curve features such as common asymptotes or the effect of changing the constant.

Several members of a one-parameter solution family approaching a common oblique asymptote.

Worked example

The general solution is $y=Ce^{-2x}+x-\tfrac12$ . Find the member through $(0,3)$ and state its oblique asymptote.

1.Substitute $x=0$ , $y=3$ : $3=C-\dfrac12$ .
2.Therefore $C=\dfrac72$ .
3.As $x\to\infty$ , $Ce^{-2x}\to0$ , so the curve approaches $y=x-\dfrac12$ .

Answer: $y=\dfrac72e^{-2x}+x-\dfrac12$ , with oblique asymptote $y=x-\dfrac12$ .

Common mistakes

Don't call a complementary-function-plus-particular-integral expression a particular solution before applying conditions.
Don't apply a derivative condition to $y$ rather than first differentiating the full general solution.
Don't sketch every family member through the same point despite the arbitrary constant changing the intercept.

Exam tip

For a family-sketch instruction, identify invariant features and show how the arbitrary constant changes the curves.

Tier 1 · Easy

1.
Find the general solution of $\dfrac{dy}{dx}=4x^3$ , then find the particular solution for which $y(1)=3$ .
(3)
(Total for Question 1 is 3 marks)
2.
Given $y=Ce^{-x}+2$ , sketch the members $C=2$ , $0$ , $-2$ on one diagram, labelling the common asymptote $y=2$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Solve $y''-y'=0$ subject to $y(0)=2$ and $y'(0)=-3$ .
(5)
(Total for Question 1 is 5 marks)
2.
Solve $y''+y=0$ subject to the boundary conditions $y(0)=1$ and $y(\pi/2)=3$ .
(5)
(Total for Question 2 is 5 marks)
3.
A solution of $y''+49y=0$ satisfies $y(0)=3$ and has a stationary point at $x=\pi/28$ . Find the particular solution, the value at the stationary point and its nature.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Find the particular solution of $y''+4y=12x$ for which $y(0)=1$ and $y'(0)=0$ .
(7)
(Total for Question 1 is 7 marks)
2.
The equation $y''-4y=\lambda x$ has a solution satisfying $y(0)=1$ , $y'(0)=0$ and $y(\ln2)=4-2\ln2$ . Determine $\lambda$ and the particular solution.
(8)
(Total for Question 2 is 8 marks)
3.
A solution of $y''+y'-30y=0$ satisfies $y(0)=1$ and $y(\ln2)=4095/64$ . Find the particular solution and the exact value of $y'(0)$ .
(8)
(Total for Question 3 is 8 marks)
4.
Find the general solution of $y'+y=2x$ . Sketch the members for which $C=1$ , $C=0$ and $C=-1$ on one diagram, and find the member whose gradient at $x=0$ is $3$ .
(8)
(Total for Question 4 is 8 marks)
5.
Find the particular solution of $y''=12x-4$ for which $y(0)=2$ and $y'(1)=0$ . Find both stationary points and determine their nature, and find all its intercepts with the $x$ -axis.
(8)
(Total for Question 5 is 8 marks)

CP-9.3 · Use differential equations in modelling in kinematics and in other contexts.

Explanation

A differential-equation model begins by defining the independent variable, dependent variable, units and any positive direction. Translate each verbal rate separately before combining them.
In kinematics, $v=ds/dt$ and $a=dv/dt$ ; a resistance force acts opposite to the velocity, so its sign follows the chosen direction.
In mixing, the rate of change is input rate minus output rate, with concentration equal to amount divided by volume.
After solving, constants, equilibrium values and limiting behaviour must be interpreted in context.
Examiners also expect modelling assumptions to be recognised, such as constant coefficients, perfect mixing or a resistance law valid over the stated range, and any non-physical prediction to be identified.

Worked example

A particle falls from rest and satisfies $\dfrac{dv}{dt}=12-0.3v$ , with downward positive. Find $v(t)$ and the terminal speed.

1.Rewrite as $v'+0.3v=12$ ; the equilibrium value satisfies $0.3v=12$ , so it is $40$ .
2.The general solution is $v=40+Ce^{-0.3t}$ .
3. $v(0)=0$ gives $C=-40$ .
4.As $t\to\infty$ , the exponential term tends to zero.

Answer: $v=40(1-e^{-0.3t})$ , and the terminal speed is $40$ in the model's speed units.

Common mistakes

Don't assign the resistance term the same sign as the velocity under the chosen positive direction.
Don't use the incoming concentration for both the input and output rates in a mixing model.
Don't report a limiting value without interpreting it as an equilibrium or terminal quantity.

Exam tip

A modelling answer should define signs and units before forming the equation and interpret the limiting value afterwards.

Tier 1 · Easy

1.
A body at temperature $T$ is in a room maintained at $18^{\circ}\text{C}$ . State a differential equation expressing that its cooling rate is proportional to its excess temperature, where $k>0$ .
(2)
(Total for Question 1 is 2 marks)
2.
A battery holds $Q$ coulombs. A charger supplies charge at $0.1(50-Q)$ coulombs per second while a device draws $2$ coulombs per second. Form a differential equation for $Q$ and find its equilibrium charge. State one modelling assumption.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
A particle falls from rest. Taking downward as positive, its speed satisfies $\dfrac{dv}{dt}=10-0.5v$ . Find $v(t)$ and the terminal speed predicted by the model.
(5)
(Total for Question 1 is 5 marks)
2.
A debt of $D$ pounds grows by continuous interest at $6\%$ per year. A further £1200 is borrowed and £3600 is repaid continuously each year. Initially $D=30000$ . Form and solve a model for $D$ , find when it predicts that the debt is cleared, and explain why the solution should not be used after that time.
(8)
(Total for Question 2 is 8 marks)
3.
A support team has $N(t)$ unresolved jobs, where $t$ is measured in hours. New jobs arrive at a constant rate of $12$ per hour, while jobs are completed at a rate of $0.3N$ per hour. Initially $N=10$ . Form and solve a differential equation, find when $N=31$ , and state the long-term value predicted by the model.
(8)
(Total for Question 3 is 8 marks)

Tier 3 · Hard

1.
A well-mixed bioreactor contains $120$ litres of liquid and initially $12$ grams of dissolved nutrient. Solution enters and leaves at $3$ litres per minute, keeping the volume constant. The incoming concentration is $0.5$ grams per litre. Form and solve a differential equation for the nutrient mass $M$ grams, find when $M=36$ , and state one modelling assumption.
(9)
(Total for Question 1 is 9 marks)
2.
A warehouse holds $S(t)$ items. Deliveries arrive at $20+2t$ items per day, while sales occur at a rate of $0.1S$ items per day. Initially $S=400$ . Form and solve the differential equation, find the exact time and amount at the minimum stock level, and comment on the long-term prediction.
(10)
(Total for Question 2 is 10 marks)
3.
There are $200$ students in a school. The number $A(t)$ who know an announcement is modelled by a rate proportional to the number who do not yet know it. Initially $20$ students know it, and after $2$ hours $110$ know it. Form and solve the model, find the number predicted to know after $6$ hours, and comment on the model.
(9)
(Total for Question 3 is 9 marks)
4.
The surrounding temperature is $(20+5\cos t)^{\circ}\text{C}$ at time $t$ hours. The temperature $T^{\circ}\text{C}$ of a body changes at half the difference between the surrounding temperature and $T$ , per hour. Given $T(0)=21$ , form and solve a differential equation. Find the minimum temperature and the first time it occurs for $t\geq0$ , and interpret the long-term temperature difference.
(10)
(Total for Question 4 is 10 marks)
5.
A drug enters a patient's bloodstream at $20e^{-t/2}$ mg h $^{-1}$ , where $t$ is measured in hours. It is removed at a rate equal to half the amount $A$ mg present, per hour, and $A(0)=0$ . Form and solve a differential equation for $A$ . Find the greatest amount and when it occurs, the total amount administered, and the long-term prediction.
(10)
(Total for Question 5 is 10 marks)

CP-9.4 · Solve differential equations of form y'' + ay' + by = 0 where a and b are constants by using the auxiliary equation.

Explanation

For the homogeneous constant-coefficient equation $y''+ay'+by=0$ , substitute $y=e^{mx}$ to form the auxiliary equation $m^2+am+b=0$ .
Its two roots determine two independent solution terms.
Distinct real roots $r,s$ give $Ae^{rx}+Be^{sx}$ ; a repeated root $r$ gives $(A+Bx)e^{rx}$ ; conjugate roots $p\pm iq$ give $e^{px}(A\cos qx+B\sin qx)$ .
Initial conditions are imposed only after the complete real general solution and any needed derivatives have been written.
Examiners expect the auxiliary equation, its roots and the matching form of solution, rather than unsupported answers or unresolved complex exponentials.

Worked example

Solve $y''-2y'+10y=0$ subject to $y(0)=1$ and $y'(0)=5$ .

1.The auxiliary equation $m^2-2m+10=0$ has roots $1\pm3i$ .
2.Thus $y=e^x(A\cos3x+B\sin3x)$ .
3. $y(0)=1$ gives $A=1$ .
4. $y'(0)=A+3B=5$ gives $B=\dfrac43$ .

Answer: $y=e^x\left(\cos3x+\dfrac43\sin3x\right)$ .

Common mistakes

Don't write the auxiliary coefficient of $m$ as $b$ instead of $a$ .
Don't use $Ae^{rx}+Be^{rx}$ for a repeated root, because this loses independence.
Don't convert roots $p\pm iq$ to trigonometric terms but omit the factor $e^{px}$ .

Exam tip

State the root type immediately after solving the auxiliary equation, then write the corresponding real solution form.

Tier 1 · Easy

1.
Find the general solution of $y''+5y'+6y=0$ .
(3)
(Total for Question 1 is 3 marks)
2.
Solve $y''-y'-6y=0$ subject to $y(0)=0$ and $y'(0)=5$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Find the general solution of $y''-6y'+9y=0$ .
(4)
(Total for Question 1 is 4 marks)
2.
Solve $4y''+4y'+y=0$ given that $y(0)=1$ and $y'(0)=0$ .
(5)
(Total for Question 2 is 5 marks)
3.
Solve $y''+4y'+3y=0$ subject to $y(0)=9$ and $y'(0)=0$ . Hence find the least value of $x>0$ for which $y''=0$ .
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Solve $2y''+4y'+10y=0$ subject to $y(0)=2$ and $y'(0)=2$ .
(6)
(Total for Question 1 is 6 marks)
2.
Solve $y''+2y'+26y=0$ subject to $y(0)=0$ and $y'(0)=5$ . Find the least positive value of $x$ at which $y=0$ .
(8)
(Total for Question 2 is 8 marks)
3.
Solve $y''-4y'+20y=0$ subject to $y(0)=1$ and $y'(0)=4$ . Find the least positive stationary point and determine its nature.
(8)
(Total for Question 3 is 8 marks)
4.
Solve $y''+10y'+34y=0$ subject to $y(0)=2$ and $y(\pi/6)=0$ . Hence find $y'(0)$ .
(8)
(Total for Question 4 is 8 marks)
5.
Solve $y''+8y'+16y=0$ subject to $y(0)=0$ and $y'(0)=1$ . Find the point of inflection of the solution curve.
(8)
(Total for Question 5 is 8 marks)

CP-9.5 · Solve differential equations of form y'' + ay' + by = f(x) by solving the homogeneous case and adding a particular integral to the complementary function (f(x) polynomial, exponential or trigonometric).

Explanation

For $y''+ay'+by=f(x)$ , the general solution is the complementary function plus a particular integral: $y=y_{\mathrm c}+y_{\mathrm p}$ . The specification uses forcing terms $ke^{px}$ , linear or quadratic polynomials, and $m\cos\omega x+n\sin\omega x$ .
Choose a trial from the same family with enough coefficients to reproduce every term after differentiation.
If that trial overlaps the complementary function, multiply the whole trial by $x$ until it is independent; a repeated overlap may require $x^2$ .
Substitute to determine the coefficients, combine CF and PI, then apply conditions.
Examiners expect resonance to be handled explicitly and conditions to be applied to the complete solution.

Worked example

Find the general solution of $y''-4y'+4y=6e^{2x}+3x$ .

1.The auxiliary equation $(m-2)^2=0$ gives $y_{\mathrm c}=(A+Bx)e^{2x}$ .
2.Because $e^{2x}$ corresponds to a repeated root, try $y_{p1}=Cx^2e^{2x}$ ; substitution gives $2Ce^{2x}=6e^{2x}$ , so $C=3$ .
3.For $3x$ , try $y_{p2}=px+q$ ; substitution gives $4px+(4q-4p)=3x$ , so $p=\dfrac34$ , $q=\dfrac34$ .

Answer: $y=(A+Bx)e^{2x}+3x^2e^{2x}+\dfrac34x+\dfrac34$ .

Common mistakes

Don't use $Ce^{2x}$ as the trial when $e^{2x}$ is already a repeated complementary-function term.
Don't use a constant trial for a linear forcing term and cannot match the coefficient of $x$ .
Don't apply initial conditions before adding the particular integral to the complementary function.

Exam tip

Before choosing a PI trial, compare every forcing term with the auxiliary roots and mark any overlap.

Tier 1 · Easy

1.
Find the general solution of $y''-y'-2y=6$ .
(4)
(Total for Question 1 is 4 marks)
2.
Find the general solution of $y''+y=2x+3$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Find the general solution of $y''-4y=8e^{2x}$ .
(5)
(Total for Question 1 is 5 marks)
2.
Find the complementary function, a particular integral and hence the general solution of $y''-y=4x^2-2$ .
(6)
(Total for Question 2 is 6 marks)
3.
Find the complementary function, a particular integral and the general solution of $y''+y'+3y=\cos2x-3\sin2x$ .
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Solve $y''+2y'+5y=10\cos x$ subject to $y(0)=1$ and $y'(0)=0$ .
(9)
(Total for Question 1 is 9 marks)
2.
Find the complementary function of $y''-2y'=4x+2$ . Explain why $y_{\mathrm p}=ax+b$ cannot be a particular integral, and hence find the solution with $y(0)=1$ , $y'(0)=0$ .
(8)
(Total for Question 2 is 8 marks)
3.
Solve $y''+2y'+2y=10\cos2x$ subject to $y(0)=0$ and $y'(0)=5$ . State the limiting form of the solution as $x\to\infty$ .
(9)
(Total for Question 3 is 9 marks)
4.
For $y''+3y'+2y=a\cos x+b\sin x$ , a particular integral is $2\cos x-\sin x$ . Find the values of $a$ and $b$ , then find the solution satisfying $y(0)=0$ and $y'(0)=0$ .
(9)
(Total for Question 4 is 9 marks)
5.
Solve $y''+4y'+4y=8x^2$ subject to $y(0)=0$ and $y'(0)=2$ .
(9)
(Total for Question 5 is 9 marks)

CP-9.6 · Understand and use the relationship between the cases when the discriminant of the auxiliary equation is positive, zero and negative and the form of solution of the differential equation.

Explanation

For the auxiliary equation $m^2+am+b=0$ , the discriminant $\Delta=a^2-4b$ determines the form of the differential-equation solution. If $\Delta>0$ , two distinct real roots give two exponential terms.
If $\Delta=0$ , the repeated root $r$ gives $(A+Bx)e^{rx}$ . If $\Delta<0$ , roots $p\pm iq$ give $e^{px}(A\cos qx+B\sin qx)$ .
The imaginary part $q$ produces oscillation; the real part $p$ controls exponential growth or decay.
Examiners expect parameter ranges to follow from a correctly simplified discriminant and the transition case to use the repeated-root form.
A negative real root alone does not imply oscillation.

Worked example

Classify the solutions of $y''+2ky'+9y=0$ for $k>0$ and write the solution when the damping is critical.

1.The auxiliary discriminant is $(2k)^2-36=4(k^2-9)$ .
2.It is negative for $0<k<3$ , zero for $k=3$ , and positive for $k>3$ .
3.At $k=3$ , the repeated auxiliary root is $-3$ .

Answer: The roots are complex for $0<k<3$ , repeated for $k=3$ , and distinct real for $k>3$ ; at $k=3$ , $y=(A+Bx)e^{-3x}$ .

Common mistakes

Don't use $a^2+4b$ instead of $a^2-4b$ for the auxiliary discriminant.
Don't write two identical exponential terms when the discriminant is zero.
Don't say that any negative auxiliary root makes the solution oscillatory.

Exam tip

For a parameter classification, solve the discriminant inequalities and state the equality case separately.

Tier 1 · Easy

1.
For $y''+4y'+8y=0$ , state the sign of the auxiliary discriminant and hence write the general solution.
(3)
(Total for Question 1 is 3 marks)
2.
State the sign of the discriminant of the auxiliary equation of $y''-3y'+2y=0$ and hence find its general solution.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Given $k>0$ , find the value of $k$ for which $y''+ky'+16y=0$ has a repeated auxiliary root. Write the corresponding general solution.
(4)
(Total for Question 1 is 4 marks)
2.
The general solution has the form $y=e^{px}(A\cos qx+B\sin qx)$ , where $q>0$ . Its trigonometric factor has period $2\pi/5$ , and its exponential factor takes the value $e^{-6}$ when $x=3$ . Determine the auxiliary roots and the differential equation in monic form.
(6)
(Total for Question 2 is 6 marks)
3.
Find the set of values of $c$ for which $y''+7y'+cy=0$ has two distinct negative real auxiliary roots.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Classify the auxiliary roots of $y''+2(p+1)y'+(p^2+4p+8)y=0$ for all real $p$ . Also give the general solution at the transition value.
(6)
(Total for Question 1 is 6 marks)
2.
Determine all real $a$ for which every non-zero solution of $y''+ay'+(2a-3)y=0$ tends to zero without oscillating. State what happens to the auxiliary roots at the boundary values within your answer.
(7)
(Total for Question 2 is 7 marks)
3.
Find the set of values of $a$ for which the period of oscillation of solutions to $y''+2ay'+(a+5)y=0$ is less than $\pi$ .
(7)
(Total for Question 3 is 7 marks)
4.
Find the set of values of $k$ for which every non-zero solution of $y''+2(k-1)y'+(k^2+1)y=0$ oscillates with a growing envelope. Describe the auxiliary roots at both boundary values.
(7)
(Total for Question 4 is 7 marks)

CP-9.7 · Solve the equation for simple harmonic motion x'' = -omega^2 x and relate the solution to the motion.

Explanation

Simple harmonic motion about equilibrium satisfies $x''=-\omega^2x$ : acceleration is proportional to displacement and directed towards equilibrium.
Its general solution is $x=A\cos\omega t+B\sin\omega t$ , equivalently $x=R\cos(\omega t+\phi)$ .
The amplitude is $R=\sqrt{A^2+B^2}$ , the period is $T=2\pi/\omega$ , maximum speed is $\omega R$ , and maximum acceleration magnitude is $\omega^2R$ .
At equilibrium the speed magnitude is greatest and acceleration is zero; at an extreme position the speed is zero and acceleration magnitude is greatest.
Examiners expect initial conditions to determine phase and amplitude and signs to be interpreted using the stated positive direction.

Displacement in simple harmonic motion, showing amplitude, equilibrium crossings and one period.

Worked example

A particle satisfies $x''=-25x$ , with $x(0)=3$ and $x'(0)=-20$ . Find its displacement, amplitude, period and maximum speed.

1. $\omega=5$ , so $x=A\cos5t+B\sin5t$ .
2. $x(0)=3$ gives $A=3$ , while $x'(0)=5B=-20$ gives $B=-4$ .
3. $R=\sqrt{3^2+(-4)^2}=5$ , $T=2\pi/5$ , and $v_{\max}=\omega R=25$ .

Answer: $x=3\cos5t-4\sin5t$ , amplitude $5$ , period $2\pi/5$ , maximum speed $25$ .

Common mistakes

Don't use $\omega=25$ after comparing $x''=-25x$ with $x''=-\omega^2x$ .
Don't calculate amplitude as $A+B$ instead of $\sqrt{A^2+B^2}$ .
Don't place maximum acceleration at equilibrium rather than at the extreme positions.

Exam tip

When asked to relate the solution to motion, state where speed and acceleration attain their extreme values.

Tier 1 · Easy

1.
A particle satisfies $x''=-9x$ . Write its general displacement and state its period.
(3)
(Total for Question 1 is 3 marks)
2.
A particle moves with $x''=-25x$ and passes through the equilibrium position with speed $15$ metres per second. Find the amplitude and the maximum magnitude of the acceleration.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
An SHM particle satisfies $x''=-4x$ , with $x(0)=4$ and $x'(0)=6$ . Find its displacement, amplitude, period and maximum speed.
(6)
(Total for Question 1 is 6 marks)
2.
At $t=0$ an SHM particle passes through equilibrium in the positive direction. It first reaches its maximum displacement of $5$ metres at $t=\pi/12$ seconds. Find $x(t)$ , its differential equation and its maximum acceleration magnitude.
(6)
(Total for Question 2 is 6 marks)
3.
An SHM particle has amplitude $4$ metres and period $2\pi/5$ seconds. At $t=0$ its displacement is $2$ metres and it is moving in the positive direction. Find $x(t)$ , its differential equation, its initial velocity and the first time it reaches displacement $-2$ metres.
(7)
(Total for Question 3 is 7 marks)

Tier 3 · Hard

1.
A particle in SHM satisfies $x''=-16x$ , $x(0)=\sqrt3/2$ and $x'(0)=-2$ . Express $x$ in the form $R\cos(4t+\phi)$ , find the first positive time at which it crosses equilibrium moving in the negative direction, and find its acceleration when $x=-1/2$ .
(8)
(Total for Question 1 is 8 marks)
2.
A particle moves with $x=6\sin(4t-\pi/3)$ , where $x$ is in metres and $t$ is in seconds. State the amplitude and period, find the velocity at $t=0$ , and find the total time in one complete oscillation for which $x>3$ .
(8)
(Total for Question 2 is 8 marks)
3.
A particle in SHM crosses equilibrium at $t=1$ moving in the negative direction and next crosses equilibrium at $t=3$ moving in the positive direction. At $t=2$ its displacement is $-7$ metres. Find $x(t)$ , the differential equation, the maximum speed and the acceleration at $t=2$ .
(8)
(Total for Question 3 is 8 marks)
4.
A particle in SHM has speed $8$ m s $^{-1}$ at equilibrium and speed $6$ m s $^{-1}$ when its displacement is $2$ m. Find the angular frequency, the amplitude and the differential equation. When at displacement $2$ m it is moving in the positive direction; find the time until it next reaches equilibrium.
(9)
(Total for Question 4 is 9 marks)

CP-9.8 · Model damped oscillations using second order differential equations and interpret their solutions.

Explanation

A free damped oscillator can be modelled by $x''+2\lambda x'+\omega_0^2x=0$ with $\lambda>0$ , where resistance is proportional to the derivative of displacement. Complex roots give underdamped motion: oscillations within a decaying exponential envelope.
A repeated negative root gives critical damping, the fastest non-oscillatory return to equilibrium.
Distinct negative real roots give overdamping and a slower non-oscillatory return.
A periodic driving force on the right produces forced vibration; the complementary function is the transient response and the particular integral is the persistent steady response.
Examiners expect classification from the auxiliary roots and an interpretation of decay, oscillation, frequency and long-term motion.

An underdamped displacement oscillating inside a decaying exponential envelope.

Worked example

Classify $x''+8x'+25x=0$ and write the form of its general solution.

1.The auxiliary equation is $m^2+8m+25=0$ .
2.Its roots are $m=-4\pm3i$ , so the motion is underdamped.
3.The negative real part supplies the decaying envelope and the imaginary part supplies angular frequency $3$ .

Answer: $x=e^{-4t}(A\cos3t+B\sin3t)$ ; the motion oscillates with exponentially decreasing amplitude.

Common mistakes

Don't classify complex roots as overdamping because their real part is negative.
Don't interpret the complementary function as the complete long-term motion in a forced-vibration equation.
Don't use displacement rather than velocity in the linear resistance term.

Exam tip

For an interpret instruction, connect the real root part to decay and the imaginary part to oscillation.

Tier 1 · Easy

1.
Classify the motion governed by $x''+6x'+25x=0$ and write its general solution.
(4)
(Total for Question 1 is 4 marks)
2.
Classify the motion modelled by $x''+7x'+10x=0$ , and explain its long-term behaviour from the auxiliary roots.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Find the positive value of $c$ for which $x''+cx'+9x=0$ is critically damped. For this value, solve the equation when $x(0)=2$ and $x'(0)=-4$ , and interpret the long-term motion.
(7)
(Total for Question 1 is 7 marks)
2.
A damped system satisfies $x''+4x'+13x=0$ , with $x(0)=3$ and $x'(0)=-6$ . Find $x(t)$ and state the damping type with a reason.
(7)
(Total for Question 2 is 7 marks)
3.
A critically damped system satisfies $x''+18x'+81x=0$ , with $x(0)=0$ and $x'(0)=9$ . Find $x(t)$ , its greatest displacement for $t\geq0$ , and describe its long-term motion.
(7)
(Total for Question 3 is 7 marks)

Tier 3 · Hard

1.
A forced damped oscillator satisfies $x''+4x'+20x=10\cos(2t)$ , with $x(0)=0$ and $x'(0)=0$ . Find $x(t)$ and describe the long-term motion.
(10)
(Total for Question 1 is 10 marks)
2.
An overdamped displacement satisfies $x''+5x'+4x=0$ , $x(0)=1$ and $x'(0)=2$ . Find $x(t)$ , its greatest value for $t\geq0$ , and prove that it approaches equilibrium without crossing it.
(9)
(Total for Question 2 is 9 marks)
3.
An overdamped displacement satisfies $x''+20x'+96x=0$ , with $x(0)=-1$ and $x'(0)=20$ . Find $x(t)$ , the first time it crosses equilibrium, its greatest displacement for $t\geq0$ , and describe its long-term motion.
(10)
(Total for Question 3 is 10 marks)
4.
A critically damped motion has displacement $x$ metres at time $t$ seconds. Its repeated auxiliary root is $-5$ , with $x(0)=3$ , and it crosses equilibrium at $t=1$ . Determine the differential equation and $x(t)$ . Find the post-crossing minimum and describe the subsequent motion.
(10)
(Total for Question 4 is 10 marks)
5.
An underdamped displacement satisfies $x''+6x'+18x=0$ , where $x$ is measured in metres and $t$ in seconds, with $x(0)=0$ and $x'(0)=6$ . Find $x(t)$ , the first positive maximum, the first positive time it next crosses equilibrium, the ratio of successive positive maxima, and its long-term behaviour.
(10)
(Total for Question 5 is 10 marks)

CP-9.9 · Analyse and interpret models with one independent variable and two dependent variables as a pair of coupled first order simultaneous equations and solve them, e.g. predator-prey models.

Explanation

The specified coupled models are linear first-order systems with one independent variable and two dependent variables, of the form $x'=ax+by+f(t)$ and $y'=cx+dy+g(t)$ . One method differentiates an equation, eliminates the other dependent variable and solves the resulting second-order equation; the eliminated variable is then recovered from an original equation.
Initial data may also determine a derivative value through the system.
Interaction terms must carry consistent signs: a transfer leaving one compartment may enter another.
Examiners expect both final functions to satisfy both original equations and the model to be interpreted through equilibria, signs and limiting behaviour.
Predictions such as negative populations mark a limit of the model's validity.

Worked example

Solve $x'=3x+y$ , $y'=x+3y$ with $x(0)=2$ and $y(0)=0$ .

1.From $y=x'-3x$ , differentiate and substitute into $y'=x+3y$ to get $x''-6x'+8x=0$ .
2.The auxiliary roots are $2$ and $4$ , so $x=Ae^{2t}+Be^{4t}$ .
3. $y=x'-3x=-Ae^{2t}+Be^{4t}$ .
4. $A+B=2$ and $-A+B=0$ , giving $A=B=1$ .

Answer: $x=e^{2t}+e^{4t}$ and $y=-e^{2t}+e^{4t}$ .

Common mistakes

Don't eliminate a variable but forget to recover it after solving the second-order equation.
Don't use only one initial condition even though the coupled system supplies two.
Don't change the sign of a transfer term in one equation without making the corresponding contextual interpretation.

Exam tip

Substitute the final pair into both first-order equations; this catches elimination and sign errors quickly.

Tier 1 · Easy

1.
Given $u'=2u+v$ and $v'=3u$ , eliminate $v$ to obtain a second-order differential equation for $u$ .
(3)
(Total for Question 1 is 3 marks)
2.
The quantities $p$ and $q$ satisfy $p'=-2p$ and $q'=p-q$ , with $p(0)=3$ . Show that $q'+q=3e^{-2t}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Solve $x'=x+2y$ , $y'=2x+y$ subject to $x(0)=3$ and $y(0)=1$ .
(7)
(Total for Question 1 is 7 marks)
2.
The quantities $x$ and $y$ satisfy $x'=-x-5y$ and $y'=x+3y$ , with $x(0)=5$ and $y(0)=0$ . Show that $x''-2x'+2x=0$ , and hence find $x$ and $y$ .
(8)
(Total for Question 2 is 8 marks)
3.
The quantities $x$ and $y$ satisfy $x'=x+5y$ and $y'=12x-6y$ , with $x(0)=2$ and $y(0)=1$ . Find $x(t)$ and $y(t)$ , and determine the limiting value of $y/x$ as $t\to\infty$ .
(8)
(Total for Question 3 is 8 marks)

Tier 3 · Hard

1.
Amounts $A(t)$ and $B(t)$ in two connected compartments are modelled by $A'=-2A+B$ and $B'=2A-3B$ , with $A(0)=4$ and $B(0)=1$ . Find $A(t)$ and $B(t)$ , determine exactly when $B$ is greatest, and state the long-term prediction.
(10)
(Total for Question 1 is 10 marks)
2.
Quantities $x(t)$ and $y(t)$ satisfy $x'=3x-2y+4$ and $y'=2x-2y$ , with $x(0)=1$ and $y(0)=3$ . Show that $x''-x'-2x=8$ . Hence find $x$ and $y$ .
(11)
(Total for Question 2 is 11 marks)
3.
Quantities $x(t)$ and $y(t)$ satisfy $x'=-8x+y+29$ and $y'=2x-9y+19$ , with $x(0)=1$ and $y(0)=3$ . Find $x$ and $y$ , find the least value of $y$ for $t\geq0$ , prove both quantities remain positive, and state their long-term values.
(11)
(Total for Question 3 is 11 marks)
4.
Quantities $x(t)$ and $y(t)$ satisfy $x'=-2x+2y$ and $y'=4x$ , with $x(0)=3$ and $y(0)=0$ . Find $x$ and $y$ , find the least value of $x$ for $t\geq0$ , and find $\lim_{t\to\infty}y/x$ .
(10)
(Total for Question 4 is 10 marks)
5.
Quantities $x(t)$ and $y(t)$ satisfy $x'=-3x+y$ and $y'=9x-3y$ , with $x(0)=3$ and $y(0)=2$ . Show that $3x+y$ is constant. Find $x$ and $y$ , find when $x=y$ , and state the limiting values.
(11)
(Total for Question 5 is 11 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CP-9.1 · Find and use an integrating factor to solve differential equations of form dy/dx + P(x)y = Q(x) and recognise when it is appropriate to do so.

Tier 1 · Easy

Mark scheme for CP-9.1 Tier 1 · Easy
Question	Scheme	Marks
1	$e^{3x}$	2
(2 marks)		2
Notes
Here $P(x)=3$ , so $I(x)=e^{\int3\,dx}=e^{3x}$ . A non-zero constant multiple would be equivalent.
2	The integrating factor is $e^{2x}$ . Multiplication gives $(e^{2x}y)'=1$ . $e^{2x}y=x+C$ . $y=(x+C)e^{-2x}$ .	4
(4 marks)		4
Notes
The integrating factor is $e^{\int2\,dx}=e^{2x}$ . Multiplying the equation by it gives $(e^{2x}y)'=1$ . Integration gives $e^{2x}y=x+C$ , and hence $y=(x+C)e^{-2x}$ .

Tier 2 · Standard

Mark scheme for CP-9.1 Tier 2 · Standard
Question	Scheme	Marks
1	$y=\dfrac{x^3}{5}+\dfrac{C}{x^2}$	5
(5 marks)		5
Notes
The integrating factor is $e^{\int2/x\,dx}=x^2$ because $x>0$ . Multiplication gives $(x^2y)'=x^4$ . Hence $x^2y=x^5/5+C$ , and division by $x^2$ gives $y=x^3/5+C/x^2$ .
2	Since the integrating factor is $e^{x^2}$ , $P(x)=(x^2)'=2x$ . $e^{x^2}y=x^3+2$ , so $(e^{x^2}y)'=3x^2$ . Hence the equation is $y'+2xy=3x^2e^{-x^2}$ . Direct substitution gives $y'+2xy=e^{-x^2}[3x^2-2x(x^3+2)+2x(x^3+2)]=3x^2e^{-x^2}$ .	4
(4 marks)		4
Notes
Since the integrating factor is $e^{x^2}$ , $P(x)=(x^2)'=2x$ . Also $e^{x^2}y=x^3+2$ , so differentiation gives $(e^{x^2}y)'=3x^2$ and therefore $y'+2xy=3x^2e^{-x^2}$ . Directly, $y'=e^{-x^2}[3x^2-2x(x^3+2)]$ ; adding $2xy=2xe^{-x^2}(x^3+2)$ leaves $3x^2e^{-x^2}$ , as required.
3	An integrating factor is $I=e^{\int\cos x\,dx}=e^{\sin x}$ . Multiplication gives $(e^{\sin x}y)'=e^{\sin x}\cos x$ . $e^{\sin x}y=e^{\sin x}+C$ , so $y=1+Ce^{-\sin x}$ . $y(0)=3$ gives $C=2$ , hence $y=1+2e^{-\sin x}$ . $y(\pi/2)=1+2/e$ .	5
(5 marks)		5
Notes
The coefficient of $y$ is $\cos x$ , so the integrating factor is $e^{\sin x}$ . The right side after multiplication is the derivative of $e^{\sin x}$ , giving $e^{\sin x}y=e^{\sin x}+C$ . Thus $y=1+Ce^{-\sin x}$ , and the initial condition gives $C=2$ . Substitution of $x=\pi/2$ gives $1+2/e$ .

Tier 3 · Hard

Mark scheme for CP-9.1 Tier 3 · Hard
Question	Scheme	Marks
1	$y=\cos x+\cos^2x$ $y(\pi/3)=\dfrac34$	7
(7 marks)		7
Notes
The integrating factor is $e^{\int2\tan x\,dx}=e^{-2\ln(\cos x)}=\sec^2x$ . Thus $(y\sec^2x)'=\sin x\sec^2x$ . Integration gives $y\sec^2x=\sec x+C$ , so $y=\cos x+C\cos^2x$ . The initial condition gives $C=1$ . At $x=\pi/3$ , $y=1/2+1/4=3/4$ .
2	$I=e^{\int 2x/(1+x^2)\,dx}=1+x^2$ . $((1+x^2)y)'=x/(1+x^2)$ . $(1+x^2)y=\tfrac12\ln(1+x^2)+C$ . $y(1)=1$ gives $C=2-\tfrac12\ln2$ . $y=\dfrac{\tfrac12\ln(1+x^2)+2-\tfrac12\ln2}{1+x^2}$ . At $x=\sqrt3$ , $1+x^2=4$ and $\tfrac12\ln4=\ln2$ . $y(\sqrt3)=\dfrac12+\dfrac{\ln2}{8}$ .	7
(7 marks)		7
Notes
On $\mathbb R$ , an integrating factor is $e^{\int2x/(1+x^2)\,dx}=1+x^2$ . Multiplication gives $((1+x^2)y)'=x/(1+x^2)$ . Hence $(1+x^2)y=\tfrac12\ln(1+x^2)+C$ . The value $y(1)=1$ gives $C=2-\tfrac12\ln2$ . At $x=\sqrt3$ , the denominator is $4$ and $\tfrac12\ln(1+x^2)=\ln2$ , so $y=1/2+(\ln2)/8$ .
3	Division by $x$ gives $y'+(1+1/x)y=xe^{-x}$ . An integrating factor is $e^{x+\ln x}=xe^x$ . $(xe^xy)'=x^2$ . $xe^xy=x^3/3+C$ . $y(1)=4/(3e)$ gives $C=1$ , so $y=(x^3+3)/(3xe^x)$ . $y(2)=11/(6e^2)$ . $y\to0$ as $x\to\infty$ .	7
(7 marks)		7
Notes
After division by $x$ , the coefficient of $y$ is $1+1/x$ , so the integrating factor on $x>0$ is $xe^x$ . The product equation is $(xe^xy)'=x^2$ , giving $xe^xy=x^3/3+C$ . At $x=1$ , the condition makes $1/3+C=4/3$ , hence $C=1$ . Substitution of $x=2$ gives $11/(6e^2)$ , and exponential growth in the denominator dominates the polynomial numerator in the limit.
4	An integrating factor is $x+1$ . Multiplication gives $((x+1)y)'=x^2-1$ . Hence $(x+1)y=x^3/3-x+k$ . Where the curve touches the $x$ -axis, $y=0$ and $y'=0$ . The differential equation then gives $x-1=0$ , so $x=1$ . Substitution in the solution gives $k=2/3$ . Then $y=(x-1)^2(x+2)/(3(x+1))\geq0$ for $x>-1$ , so $y$ is least at $x=1$ .	7
(7 marks)		7
Notes
The integrating factor $x+1$ converts the equation to a product derivative. The initial condition is the integration constant. Where the curve touches the axis, both $y$ and $y'$ are zero, so the original equation fixes $x$ before the solution fixes $k$ . Factoring then proves the least-value claim on the stated domain.
5	An integrating factor is $e^{\int2\tanh x\,dx}=\cosh^2x$ . Multiplication gives $(y\cosh^2x)'=3\sinh x\cosh x$ . Hence $y=3/2+C\operatorname{sech}^2x$ . $y_1=3/2+(5/2)\operatorname{sech}^2x$ and $y_2=3/2-(1/2)\operatorname{sech}^2x$ . Their separation is $3\operatorname{sech}^2x$ . The area is $3[\tanh x]_0^{\ln2}=9/5$ . Their limiting separation is $0$ .	7
(7 marks)		7
Notes
The integrating factor is $\cosh^2x$ . After integration, each solution differs only in the constant multiplying $\operatorname{sech}^2x$ . Their difference therefore integrates directly using the derivative of tanh and tends to zero with the squared hyperbolic secant.

CP-9.2 · Find both general and particular solutions to differential equations.

Tier 1 · Easy

Mark scheme for CP-9.2 Tier 1 · Easy
Question	Scheme	Marks
1	General: $y=x^4+C$ Particular: $y=x^4+2$	3
(3 marks)		3
Notes
Integrating gives $y=x^4+C$ . Substituting $x=1$ , $y=3$ gives $3=1+C$ , so $C=2$ .
2	For $C=2$ , the curve passes through $(0,4)$ , decreases and approaches $y=2$ from above as $x\to\infty$ . For $C=0$ , the member is the horizontal line $y=2$ . For $C=-2$ , the curve passes through $(0,0)$ , increases and approaches $y=2$ from below as $x\to\infty$ . The common horizontal asymptote is $y=2$ .	4
(4 marks)		4
Notes
Substitute each value of $C$ . Since $e^{-x}>0$ and decreases to $0$ as $x\to\infty$ , the $C=2$ member lies above $y=2$ and decreases, while the $C=-2$ member lies below $y=2$ and increases. At $x=0$ they have values $4$ , $2$ and $0$ , respectively; all three tend to $2$ as $x\to\infty$ .

Tier 2 · Standard

Mark scheme for CP-9.2 Tier 2 · Standard
Question	Scheme	Marks
1	$y=5-3e^x$	5
(5 marks)		5
Notes
The auxiliary equation is $m^2-m=0$ , with roots $0$ and $1$ . Hence $y=A+Be^x$ and $y'=Be^x$ . From $y'(0)=-3$ , $B=-3$ ; then $y(0)=A+B=2$ gives $A=5$ .
2	The auxiliary equation is $m^2+1=0$ . The roots are $m=\pm i$ . $y=A\cos x+B\sin x$ . $y(0)=1$ gives $A=1$ . $y(\pi/2)=3$ gives $B=3$ , so $y=\cos x+3\sin x$ .	5
(5 marks)		5
Notes
The auxiliary roots are $\pm i$ , so $y=A\cos x+B\sin x$ . The condition at $x=0$ gives $A=1$ . At $x=\pi/2$ , the cosine term is zero and the sine term is $B$ , hence $B=3$ .
3	The auxiliary equation is $m^2+49=0$ . Its roots are $m=\pm7i$ . $y=A\cos7x+B\sin7x$ . $y(0)=3$ gives $A=3$ , and $y'(\pi/28)=0$ gives $B=3$ . At $x=\pi/28$ , $y=3\sqrt2$ and $y''=-49y<0$ , so the point is a maximum.	5
(5 marks)		5
Notes
The auxiliary roots are $\pm7i$ , giving $y=A\cos7x+B\sin7x$ . The initial value sets $A=3$ . Since $y'=7(-A\sin7x+B\cos7x)$ , the stationary condition at angle $\pi/4$ gives $B=A=3$ . Substitution gives $y=3\sqrt2$ there. The differential equation gives $y''=-49y<0$ , so the stationary point is a maximum.

Tier 3 · Hard

Mark scheme for CP-9.2 Tier 3 · Hard
Question	Scheme	Marks
1	$y=\cos(2x)-\dfrac32\sin(2x)+3x$	7
(7 marks)		7
Notes
The complementary function is $A\cos2x+B\sin2x$ . A linear particular integral $px+q$ gives $4(px+q)=12x$ , so $p=3$ and $q=0$ . Thus $y=A\cos2x+B\sin2x+3x$ . The first condition gives $A=1$ . Since $y'=-2A\sin2x+2B\cos2x+3$ , the second gives $2B+3=0$ , so $B=-3/2$ .
2	The auxiliary roots are $2$ and $-2$ , so $y_{\mathrm c}=Ae^{2x}+Be^{-2x}$ . A linear particular integral is $y_{\mathrm p}=-(\lambda/4)x$ . $y=Ae^{2x}+Be^{-2x}-(\lambda/4)x$ . $y(0)=1$ and $y'(0)=0$ give $A+B=1$ and $A-B=\lambda/8$ . $A=1/2+\lambda/16$ and $B=1/2-\lambda/16$ . At $x=\ln2$ , the condition becomes $17/8+\lambda(15/64-(\ln2)/4)=4-2\ln2$ . Since $4-2\ln2-17/8=8(15/64-(\ln2)/4)$ and the bracket is non-zero, $\lambda=8$ . Then $A=1$ , $B=0$ , so the particular solution is $y=e^{2x}-2x$ .	8
(8 marks)		8
Notes
For general $\lambda$ , $y=Ae^{2x}+Be^{-2x}-(\lambda/4)x$ . The first two conditions give $A+B=1$ and $A-B=\lambda/8$ , so $A=1/2+\lambda/16$ and $B=1/2-\lambda/16$ . At $x=\ln2$ , the boundary value becomes $17/8+\lambda(15/64-(\ln2)/4)=4-2\ln2$ . The right side minus $17/8$ equals $8(15/64-(\ln2)/4)$ , and the bracket is non-zero, so $\lambda=8$ . Then $A=1$ , $B=0$ , giving $y=e^{2x}-2x$ .
3	The auxiliary equation is $m^2+m-30=0$ . Its roots are $5$ and $-6$ . $y=Ae^{5x}+Be^{-6x}$ . $y(0)=1$ gives $A+B=1$ . $y(\ln2)=4095/64$ gives $32A+B/64=4095/64$ . Solving gives $A=2$ and $B=-1$ . The particular solution is $y=2e^{5x}-e^{-6x}$ . $y'(0)=10+6=16$ .	8
(8 marks)		8
Notes
Factoring the auxiliary equation gives roots $5$ and $-6$ . The two boundary conditions give $A+B=1$ and $32A+B/64=4095/64$ . Multiplying the second equation by $64$ and using $B=1-A$ gives $2047A+1=4095$ , so $A=2$ and $B=-1$ . Differentiation of the resulting solution gives $y'=10e^{5x}+6e^{-6x}$ , hence $y'(0)=16$ .
4	An integrating factor is $e^x$ . Multiplication gives $(e^xy)'=2xe^x$ . Hence the general solution is $y=2x-2+Ce^{-x}$ . The three members are $y=2x-2+e^{-x}$ , $y=2x-2$ and $y=2x-2-e^{-x}$ . All three approach the common oblique asymptote $y=2x-2$ as $x\to\infty$ . Their $y$ -intercepts are $-1$ , $-2$ and $-3$ , respectively, which fixes their labelled order on the sketch. $y'=2-Ce^{-x}$ , so $y'(0)=2-C$ . The required gradient gives $C=-1$ , so the required member is $y=2x-2-e^{-x}$ .	8
(8 marks)		8
Notes
Use the integrating factor $e^x$ and integrate the resulting product derivative to obtain the one-parameter family. The common asymptote and the three intercepts determine the requested labelled sketches. Differentiate the family and impose the gradient at zero to select $C=-1$ .
5	One integration gives $y'=6x^2-4x+A$ . $y'(1)=0$ gives $A=-2$ . A second integration gives $y=2x^3-2x^2-2x+B$ . $y(0)=2$ gives $B=2$ . $y'=2(3x+1)(x-1)$ , so the stationary points have $x$ -coordinates $-1/3$ and $1$ . The point $(-1/3,64/27)$ is a maximum. The point $(1,0)$ is a minimum. $y=2(x-1)^2(x+1)$ , so the $x$ -intercepts are $(-1,0)$ and $(1,0)$ , with tangency at $(1,0)$ .	8
(8 marks)		8
Notes
Integrate twice and apply the derivative and value conditions to determine the two constants. Factoring the derivative locates the stationary points; $y''=12x-4$ distinguishes their nature. Factoring the completed particular solution gives both axis intersections and shows the repeated root at the minimum.

CP-9.3 · Use differential equations in modelling in kinematics and in other contexts.

Tier 1 · Easy

Mark scheme for CP-9.3 Tier 1 · Easy
Question	Scheme	Marks
1	$\dfrac{dT}{dt}=-k(T-18)$	2
(2 marks)		2
Notes
The excess temperature is $T-18$ . Cooling means $T$ decreases when this excess is positive, so the proportionality constant must appear with a minus sign.
2	The charger supplies $0.1(50-Q)=5-0.1Q$ coulombs per second. After the device draw is subtracted, $\dfrac{dQ}{dt}=3-0.1Q$ . At equilibrium $dQ/dt=0$ , so $Q=30$ coulombs. The model assumes, for example, that both stated rates remain valid throughout the modelled interval.	4
(4 marks)		4
Notes
The net rate is charger input minus device output: $Q'=0.1(50-Q)-2=3-0.1Q$ . At equilibrium $Q'=0$ , so $Q=30$ . This uses the assumption that the draw is constant and the charging law does not change with temperature, age or time.

Tier 2 · Standard

Mark scheme for CP-9.3 Tier 2 · Standard
Question	Scheme	Marks
1	$v=20(1-e^{-t/2})$ Terminal speed $20$	5
(5 marks)		5
Notes
Rewrite as $v'+0.5v=10$ . The general solution is $v=20+Ce^{-t/2}$ . Since $v(0)=0$ , $C=-20$ , giving $v=20(1-e^{-t/2})$ . As $t\to\infty$ , the exponential tends to $0$ , so the limiting speed is $20$ in the stated units.
2	Continuous interest contributes $0.06D$ pounds per year. The net fixed cash flow is $1200-3600=-2400$ pounds per year. Hence $D'=0.06D-2400$ . Solving gives $D=40000+Ce^{0.06t}$ . $D(0)=30000$ gives $C=-10000$ , so $D=40000-10000e^{0.06t}$ . Setting $D=0$ gives $e^{0.06t}=4$ . The predicted clearing time is $t=\dfrac{\ln4}{0.06}$ years. After this time the formula predicts negative debt, so the repayment rule must stop or change.	8
(8 marks)		8
Notes
Interest contributes $0.06D$ per year and the net fixed cash flow is $1200-3600=-2400$ , giving $D'-0.06D=-2400$ . The solution is $D=40000+Ce^{0.06t}$ ; $D(0)=30000$ gives $C=-10000$ . Setting $D=0$ gives $e^{0.06t}=4$ , hence $t=(\ln4)/0.06$ . Continuing the same equation would make $D<0$ , contradicting the interpretation as money owed.
3	New jobs contribute $12$ jobs per hour. Completed jobs remove $0.3N$ jobs per hour. Hence $N'=12-0.3N$ . The general solution is $N=40+Ce^{-0.3t}$ . $N(0)=10$ gives $N=40-30e^{-0.3t}$ . $N=31$ gives $e^{-0.3t}=3/10$ . The time is $t=\dfrac{10}{3}\ln(10/3)$ hours. As $t\to\infty$ , the model predicts $N\to40$ jobs.	8
(8 marks)		8
Notes
The rate of change is arrivals minus completions, giving $N'+0.3N=12$ . Its equilibrium is $40$ and the general solution is $40+Ce^{-0.3t}$ . The initial value gives $C=-30$ . Setting $N=31$ gives $30e^{-0.3t}=9$ , so $e^{-0.3t}=3/10$ and $t=(10/3)\ln(10/3)$ . The exponential term tends to zero, leaving the long-term value $40$ .

Tier 3 · Hard

Mark scheme for CP-9.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\dfrac{dM}{dt}=1.5-\dfrac{M}{40}$ $M=60-48e^{-t/40}$ $t=40\ln2$ minutes For example, the nutrient is mixed uniformly at every instant.	9
(9 marks)		9
Notes
Nutrient enters at $3(0.5)=1.5$ grams per minute. Its concentration in the tank is $M/120$ , so it leaves at $3M/120=M/40$ . Thus $M'+M/40=1.5$ , whose solution is $M=60+Ce^{-t/40}$ . The initial mass gives $C=-48$ . Setting $M=36$ gives $24=48e^{-t/40}$ , so $e^{-t/40}=1/2$ and $t=40\ln2$ . The derivation assumes instantaneous perfect mixing, as well as constant rates and volume.
2	Deliveries contribute $20+2t$ items per day and sales remove $0.1S$ items per day. Thus $S'+0.1S=20+2t$ . A linear particular integral is $20t$ , so $S=20t+Ce^{-0.1t}$ . $S(0)=400$ gives $S=20t+400e^{-0.1t}$ . $S'=20-40e^{-0.1t}$ . $S'=0$ gives $e^{-0.1t}=1/2$ , so $t=10\ln2$ . $S''=4e^{-0.1t}>0$ , so this stationary point is a minimum. The minimum stock is $S(10\ln2)=200\ln2+200$ items. As $t\to\infty$ , $S\sim20t$ , so the model predicts unbounded, asymptotically linear stock growth. Consequently fixed sales behaviour or unlimited warehouse capacity is eventually unrealistic.	10
(10 marks)		10
Notes
Net change is delivery rate minus sales rate, so $S'+0.1S=20+2t$ . A linear PI is $20t$ , and the complementary term is $Ce^{-0.1t}$ ; $S(0)=400$ gives $C=400$ . Thus $S'=20-40e^{-0.1t}$ . The stationary condition gives $e^{-0.1t}=1/2$ , hence $t=10\ln2$ . Since $S''=4e^{-0.1t}>0$ , this is a minimum, and substitution gives $200\ln2+200$ . As $t\to\infty$ , $S\sim20t$ , exposing the model limitation.
3	The number who do not know is $200-A$ , so $A'=k(200-A)$ for a constant $k>0$ . The general solution is $A=200+Ce^{-kt}$ . $A(0)=20$ gives $C=-180$ . $A(2)=110$ gives $90=180e^{-2k}$ . Hence $k=\tfrac12\ln2$ . $A=200-180e^{-(\ln2)t/2}$ . $A(6)=200-180e^{-3\ln2}=200-180/8=177.5$ . The non-integral prediction cannot be a literal student count, showing that the continuous model is an approximation. The model also ignores differences in contact patterns between students.	9
(9 marks)		9
Notes
The stated proportionality gives $A'=k(200-A)$ . Solving gives $A=200+Ce^{-kt}$ , and the initial value gives $C=-180$ . The observation at two hours gives $90=180e^{-2k}$ , so $e^{-2k}=1/2$ and $k=(\ln2)/2$ . At six hours the exponential is $e^{-3\ln2}=1/8$ , giving $A(6)=177.5$ . Since people are counted discretely, this value exposes the continuous approximation; uniform contact behaviour is another idealisation.
4	The model is $T'=\tfrac12(20+5\cos t-T)$ with $T(0)=21$ . In linear form, $T'+\tfrac12T=10+\tfrac52\cos t$ . An integrating factor is $e^{t/2}$ . For $T_{\mathrm p}=20+a\cos t+b\sin t$ , coefficient comparison gives $a=1$ and $b=2$ . Thus $T=20+\cos t+2\sin t+Ce^{-t/2}$ . $T(0)=21$ gives $C=0$ . Write $T=20+\sqrt5\cos(t-\arctan 2)$ . The first minimum occurs at $t=\pi+\arctan 2$ hours. The minimum temperature is $(20-\sqrt5)^{\circ}\text{C}$ . $T-(20+5\cos t)=2\sin t-4\cos t$ , so the difference remains periodic and has no limiting constant value.	10
(10 marks)		10
Notes
Translate the proportional-rate statement with the periodic surrounding temperature, then solve the resulting linear equation using a sine-cosine particular integral. The initial condition removes the transient term. Phase-amplitude form gives the first minimum, while direct subtraction shows that the long-term temperature difference remains periodic rather than approaching a constant.
5	Input minus removal gives $A'+A/2=20e^{-t/2}$ with $A(0)=0$ . An integrating factor is $e^{t/2}$ . Multiplication gives $(Ae^{t/2})'=20$ . Integration gives $Ae^{t/2}=20t+C$ . $A(0)=0$ gives $C=0$ . Therefore $A=20te^{-t/2}$ . $A'=20e^{-t/2}(1-t/2)$ , so the greatest amount occurs at $t=2$ hours. The greatest amount is $40/e$ mg. Evaluating $\int_0^R 20e^{-t/2}\,dt=40-40e^{-R/2}$ and letting $R\to\infty$ gives a total of $40$ mg. The model predicts $A\to0$ as $t\to\infty$ .	10
(10 marks)		10
Notes
Use input rate minus removal rate to form a linear equation. The integrating factor cancels the decreasing input exponential, leaving a constant product derivative. Differentiate the solution for its maximum, evaluate the administered amount to a finite upper limit before taking the improper limit, and take the solution's long-term limit.

CP-9.4 · Solve differential equations of form y'' + ay' + by = 0 where a and b are constants by using the auxiliary equation.

Tier 1 · Easy

Mark scheme for CP-9.4 Tier 1 · Easy
Question	Scheme	Marks
1	$y=Ae^{-2x}+Be^{-3x}$	3
(3 marks)		3
Notes
The auxiliary equation $m^2+5m+6=0$ factorises as $(m+2)(m+3)=0$ . The distinct roots are $-2$ and $-3$ , giving the stated two-exponential solution.
2	The auxiliary equation factorises as $(m-3)(m+2)=0$ , giving roots $3$ and $-2$ . $y=Ae^{3x}+Be^{-2x}$ . The conditions give $A+B=0$ and $3A-2B=5$ . Thus $A=1$ , $B=-1$ , and $y=e^{3x}-e^{-2x}$ .	4
(4 marks)		4
Notes
The auxiliary equation $(m-3)(m+2)=0$ gives $y=Ae^{3x}+Be^{-2x}$ . The conditions give $A+B=0$ and $3A-2B=5$ . Substituting $B=-A$ gives $5A=5$ , so $A=1$ and $B=-1$ .

Tier 2 · Standard

Mark scheme for CP-9.4 Tier 2 · Standard
Question	Scheme	Marks
1	$y=(A+Bx)e^{3x}$	4
(4 marks)		4
Notes
The auxiliary equation is $(m-3)^2=0$ , so $m=3$ is a repeated root. The two independent terms are $e^{3x}$ and $xe^{3x}$ .
2	The auxiliary equation is $4m^2+4m+1=(2m+1)^2=0$ . The repeated root is $m=-1/2$ . $y=(A+Bx)e^{-x/2}$ . $y(0)=1$ gives $A=1$ . $y'(0)=B-A/2=0$ gives $B=1/2$ , so $y=(1+x/2)e^{-x/2}$ .	5
(5 marks)		5
Notes
The auxiliary equation $4m^2+4m+1=(2m+1)^2=0$ has repeated root $-1/2$ , so $y=(A+Bx)e^{-x/2}$ . From $y(0)=1$ , $A=1$ . Differentiation gives $y'(0)=B-A/2=0$ , hence $B=1/2$ and $y=(1+x/2)e^{-x/2}$ .
3	The auxiliary equation $(m+1)(m+3)=0$ has roots $-1$ and $-3$ . $y=Ae^{-x}+Be^{-3x}$ . The conditions give $A+B=9$ and $-A-3B=0$ . $A=27/2$ and $B=-9/2$ , so $y=\tfrac{27}{2}e^{-x}-\tfrac92e^{-3x}$ . $y''=\tfrac{27}{2}e^{-x}-\tfrac{81}{2}e^{-3x}$ . $y''=0$ gives $e^{2x}=3$ , so the least positive value is $x=\tfrac12\ln3$ .	6
(6 marks)		6
Notes
The auxiliary roots are $-1$ and $-3$ , giving $y=Ae^{-x}+Be^{-3x}$ . The initial conditions give $A+B=9$ and $-A-3B=0$ , hence $A=27/2$ and $B=-9/2$ . Differentiating twice gives $y''=(27/2)e^{-x}-(81/2)e^{-3x}$ . Its zeros satisfy $e^{2x}=3$ , which has the single positive solution $x=(1/2)\ln3$ .

Tier 3 · Hard

Mark scheme for CP-9.4 Tier 3 · Hard
Question	Scheme	Marks
1	$y=2e^{-x}(\cos2x+\sin2x)$	6
(6 marks)		6
Notes
After division by $2$ , the auxiliary equation is $m^2+2m+5=0$ , with roots $-1\pm2i$ . Thus $y=e^{-x}(A\cos2x+B\sin2x)$ . The first condition gives $A=2$ . Differentiating and setting $x=0$ gives $y'(0)=-A+2B=2$ , so $B=2$ .
2	The auxiliary equation is $m^2+2m+26=0$ . Its roots are $-1\pm5i$ . $y=e^{-x}(A\cos5x+B\sin5x)$ . $y(0)=0$ gives $A=0$ . $y'(0)=5B=5$ gives $B=1$ . Hence $y=e^{-x}\sin5x$ . Since $e^{-x}>0$ , positive zeros occur when $5x=n\pi$ for positive integers $n$ . The least positive value is $x=\pi/5$ .	8
(8 marks)		8
Notes
The auxiliary roots are $-1\pm5i$ , so $y=e^{-x}(A\cos5x+B\sin5x)$ . The condition $y(0)=0$ gives $A=0$ , and then $y'(0)=5B=5$ gives $B=1$ . Thus $y=e^{-x}\sin5x$ . The exponential factor is always positive, so the first zero after $x=0$ occurs when $5x=\pi$ , giving $x=\pi/5$ .
3	The auxiliary equation $m^2-4m+20=0$ has roots $2\pm4i$ . $y=e^{2x}(A\cos4x+B\sin4x)$ . $y(0)=1$ gives $A=1$ . $y'(0)=4$ gives $2A+4B=4$ , so $B=1/2$ . $y=e^{2x}(\cos4x+\tfrac12\sin4x)$ . $y'=e^{2x}(4\cos4x-3\sin4x)$ . The least positive stationary value occurs at $x=\tfrac14\arctan(4/3)$ and is a maximum. Its $y$ -coordinate is $e^{\frac12\arctan(4/3)}$ .	8
(8 marks)		8
Notes
The roots $2\pm4i$ give the stated real solution. The initial conditions set $A=1$ and $B=1/2$ . Differentiation simplifies to $y'=e^{2x}(4\cos4x-3\sin4x)$ . The first positive zero therefore has $\tan4x=4/3$ , so $x=\tfrac14\arctan(4/3)$ . The derivative changes from positive to negative there. With $\cos4x=3/5$ and $\sin4x=4/5$ , the trigonometric factor in $y$ equals $1$ , giving the stated ordinate.
4	The auxiliary equation is $m^2+10m+34=0$ . Its roots are $-5\pm3i$ . $y=e^{-5x}(A\cos3x+B\sin3x)$ . $y(0)=2$ gives $A=2$ . $y(\pi/6)=0$ gives $Be^{-5\pi/6}=0$ . Hence $B=0$ and $y=2e^{-5x}\cos3x$ . $y'=2e^{-5x}(-5\cos3x-3\sin3x)$ . Therefore $y'(0)=-10$ .	8
(8 marks)		8
Notes
The fresh complex auxiliary roots give an exponentially scaled sine-cosine solution. The value at zero fixes the cosine coefficient, while the boundary value at $\pi/6$ fixes the sine coefficient because $\cos(\pi/2)=0$ . Differentiate the resulting solution at zero.
5	The auxiliary equation is $(m+4)^2=0$ . $y=(A+Bx)e^{-4x}$ . $y(0)=0$ gives $A=0$ . $y'=Be^{-4x}(1-4x)$ . $y'(0)=1$ gives $B=1$ , so $y=xe^{-4x}$ . $y''=8e^{-4x}(2x-1)$ . $y''$ changes sign at $x=1/2$ . The point of inflection is $(1/2,e^{-2}/2)$ .	8
(8 marks)		8
Notes
The repeated negative root gives a linear factor times $e^{-4x}$ . The two initial conditions determine its coefficients. The second derivative then changes sign once, giving the exact inflection point.

CP-9.5 · Solve differential equations of form y'' + ay' + by = f(x) by solving the homogeneous case and adding a particular integral to the complementary function (f(x) polynomial, exponential or trigonometric).

Tier 1 · Easy

Mark scheme for CP-9.5 Tier 1 · Easy
Question	Scheme	Marks
1	$y=Ae^{2x}+Be^{-x}-3$	4
(4 marks)		4
Notes
The auxiliary equation $(m-2)(m+1)=0$ gives $y_{\mathrm c}=Ae^{2x}+Be^{-x}$ . For a constant PI $K$ , substitution gives $-2K=6$ , so $K=-3$ .
2	The auxiliary equation $m^2+1=0$ has roots $\pm i$ . Thus $y_{\mathrm c}=A\cos x+B\sin x$ . Try $y_{\mathrm p}=ax+b$ ; substitution gives $ax+b=2x+3$ . $a=2$ , $b=3$ , so $y=A\cos x+B\sin x+2x+3$ .	4
(4 marks)		4
Notes
The auxiliary roots are $\pm i$ , giving $y_{\mathrm c}=A\cos x+B\sin x$ . For the polynomial forcing, try $y_{\mathrm p}=ax+b$ . Substitution gives $ax+b=2x+3$ , so $a=2$ and $b=3$ .

Tier 2 · Standard

Mark scheme for CP-9.5 Tier 2 · Standard
Question	Scheme	Marks
1	$y=Ae^{2x}+Be^{-2x}+2xe^{2x}$	5
(5 marks)		5
Notes
The complementary function is $Ae^{2x}+Be^{-2x}$ . Since $e^{2x}$ is already present, try $y_{\mathrm p}=Cxe^{2x}$ . Then $y_{\mathrm p}''-4y_{\mathrm p}=4Ce^{2x}$ , so $C=2$ .
2	The auxiliary equation $m^2-1=0$ has roots $1$ and $-1$ . $y_{\mathrm c}=Ae^x+Be^{-x}$ . For a particular integral, take $y_{\mathrm p}=ax^2+bx+c$ . Then $y_{\mathrm p}''-y_{\mathrm p}=-ax^2-bx+(2a-c)$ . Matching $4x^2-2$ gives $a=-4$ , $b=0$ , $c=-6$ , so $y_{\mathrm p}=-4x^2-6$ . Hence $y=Ae^x+Be^{-x}-4x^2-6$ .	6
(6 marks)		6
Notes
The auxiliary roots are $1$ and $-1$ . For a PI, take $ax^2+bx+c$ . Then $y_{\mathrm p}''-y_{\mathrm p}=-ax^2-bx+(2a-c)$ . Matching $4x^2-2$ gives $a=-4$ , $b=0$ and $c=-6$ . Add this PI to the complementary function.
3	The auxiliary equation is $m^2+m+3=0$ . Its roots are $(-1\pm i\sqrt{11})/2$ . $y_{\mathrm c}=e^{-x/2}\left(A\cos(\sqrt{11}x/2)+B\sin(\sqrt{11}x/2)\right)$ . For $y_{\mathrm p}=a\cos2x+b\sin2x$ , coefficient comparison gives $-a+2b=1$ and $-2a-b=-3$ . These equations give $a=b=1$ , so $y_{\mathrm p}=\cos2x+\sin2x$ . $y=y_{\mathrm c}+\cos2x+\sin2x$ .	6
(6 marks)		6
Notes
The auxiliary discriminant is $1-12=-11$ , giving the stated complementary function. Substitute $a\cos2x+b\sin2x$ into the left side. The cosine coefficient is $-a+2b$ and the sine coefficient is $-2a-b$ . Matching the forcing gives $a=b=1$ , after which the complementary function and particular integral are added.

Tier 3 · Hard

Mark scheme for CP-9.5 Tier 3 · Hard
Question	Scheme	Marks
1	$y=-e^{-x}(\cos2x+\sin2x)+2\cos x+\sin x$	9
(9 marks)		9
Notes
The auxiliary roots are $-1\pm2i$ , so $y_{\mathrm c}=e^{-x}(A\cos2x+B\sin2x)$ . Try $y_{\mathrm p}=a\cos x+b\sin x$ . Substitution gives $4a+2b=10$ and $-2a+4b=0$ , so $a=2$ , $b=1$ . From $y(0)=1$ , $A+2=1$ and $A=-1$ . The derivative at $0$ is $-A+2B+1$ ; setting this to $0$ gives $B=-1$ . Combining the parts yields the stated solution.
2	The auxiliary equation is $m^2-2m=m(m-2)=0$ . $y_{\mathrm c}=A+Be^{2x}$ . If $y_{\mathrm p}=ax+b$ , then $y_{\mathrm p}''-2y_{\mathrm p}'=-2a$ , which cannot reproduce the term $4x$ . The constant part also overlaps the constant term in the complementary function, so use $y_{\mathrm p}=ax^2+bx$ . Substitution gives $-4ax+(2a-2b)=4x+2$ . $a=-1$ and $b=-2$ , so $y_{\mathrm p}=-x^2-2x$ . $y=A+Be^{2x}-x^2-2x$ , and the initial conditions give $A+B=1$ , $2B-2=0$ . $A=0$ , $B=1$ , so $y=e^{2x}-x^2-2x$ .	8
(8 marks)		8
Notes
The auxiliary roots are $0$ and $2$ , so $y_{\mathrm c}=A+Be^{2x}$ . A trial $ax+b$ gives only the constant $-2a$ after substitution and cannot match $4x+2$ ; its constant part also overlaps the complementary function. Instead take $y_{\mathrm p}=ax^2+bx$ . Then $y_{\mathrm p}''-2y_{\mathrm p}'=-4ax+(2a-2b)$ , so $a=-1$ and $b=-2$ . Thus $y=A+Be^{2x}-x^2-2x$ . The initial conditions give $A+B=1$ and $2B-2=0$ , hence $B=1$ , $A=0$ and $y=e^{2x}-x^2-2x$ .
3	The auxiliary roots are $-1\pm i$ , so $y_{\mathrm c}=e^{-x}(A\cos x+B\sin x)$ . Use $y_{\mathrm p}=a\cos2x+b\sin2x$ . Substitution gives cosine and sine coefficients $-2a+4b$ and $-4a-2b$ . Matching $10\cos2x$ gives $-2a+4b=10$ and $-4a-2b=0$ , so $a=-1$ and $b=2$ . $y=e^{-x}(A\cos x+B\sin x)-\cos2x+2\sin2x$ . $y(0)=0$ gives $A=1$ . $y'(0)=5$ gives $-A+B+4=5$ , so $B=2$ . $y=e^{-x}(\cos x+2\sin x)-\cos2x+2\sin2x$ . As $x\to\infty$ , the decaying complementary terms disappear and the limiting form is $-\cos2x+2\sin2x$ .	9
(9 marks)		9
Notes
The auxiliary equation has roots $-1\pm i$ . For the trigonometric trial, substitution into $y''+2y'+2y$ gives $(-2a+4b)\cos2x+(-4a-2b)\sin2x$ . Matching $10\cos2x$ gives $a=-1$ , $b=2$ . The value condition gives $A=1$ . The complementary derivative contributes $-A+B$ at zero and the particular integral contributes $4$ , so $B=2$ . The factor $e^{-x}$ forces the complementary part to zero in the limit.
4	$y_{\mathrm p}'=-2\sin x-\cos x$ and $y_{\mathrm p}''=-2\cos x+\sin x$ . Substitution gives $-\cos x-7\sin x$ . Hence $a=-1$ and $b=-7$ . The auxiliary equation is $(m+1)(m+2)=0$ . $y=Ae^{-x}+Be^{-2x}+2\cos x-\sin x$ . $y(0)=0$ gives $A+B=-2$ . $y'(0)=0$ gives $A+2B=-1$ . Thus $A=-3$ and $B=1$ . The solution is $y=-3e^{-x}+e^{-2x}+2\cos x-\sin x$ .	9
(9 marks)		9
Notes
Differentiate the supplied particular integral and substitute it to recover the forcing coefficients. Add the complementary function from the two negative real roots, then apply the two initial conditions to determine its constants.
5	The auxiliary equation is $(m+2)^2=0$ . $y_{\mathrm c}=(A+Bx)e^{-2x}$ . Use $y_{\mathrm p}=ax^2+bx+c$ . Substitution gives $4ax^2+(8a+4b)x+(2a+4b+4c)$ . Coefficient comparison gives $a=2$ . It then gives $b=-4$ and $c=3$ . $y=(A+Bx)e^{-2x}+2x^2-4x+3$ . $y(0)=0$ gives $A=-3$ , and $y'(0)=2$ gives $B=0$ . Therefore $y=-3e^{-2x}+2x^2-4x+3$ .	9
(9 marks)		9
Notes
The repeated root supplies the complementary function. A quadratic trial matches the polynomial forcing; comparing coefficients gives $2x^2-4x+3$ . The initial value fixes $A$ , and differentiating the complete solution then fixes $B$ .

CP-9.6 · Understand and use the relationship between the cases when the discriminant of the auxiliary equation is positive, zero and negative and the form of solution of the differential equation.

Tier 1 · Easy

Mark scheme for CP-9.6 Tier 1 · Easy
Question	Scheme	Marks
1	The discriminant is negative; $y=e^{-2x}(A\cos2x+B\sin2x)$ .	3
(3 marks)		3
Notes
The discriminant is $4^2-4(8)=-16<0$ . The roots are $-2\pm2i$ , which give the damped sine-cosine form.
2	$\Delta=(-3)^2-4(1)(2)=1>0$ . The auxiliary roots are $1$ and $2$ . $y=Ae^x+Be^{2x}$ .	3
(3 marks)		3
Notes
The auxiliary equation is $m^2-3m+2=0$ . Its discriminant is $(-3)^2-4(1)(2)=1>0$ , so there are two distinct real roots. Factorising gives $(m-1)(m-2)=0$ , hence $y=Ae^x+Be^{2x}$ .

Tier 2 · Standard

Mark scheme for CP-9.6 Tier 2 · Standard
Question	Scheme	Marks
1	$k=8$ ; $y=(A+Bx)e^{-4x}$	4
(4 marks)		4
Notes
A repeated root requires discriminant $k^2-64=0$ . Since $k>0$ , $k=8$ . The repeated root is $-k/2=-4$ , so the solution is $(A+Bx)e^{-4x}$ .
2	For roots $p\pm qi$ , the trigonometric factor has period $2\pi/\|q\|$ . $2\pi/\|q\|=2\pi/5$ , so $\|q\|=5$ . $e^{3p}=e^{-6}$ . $p=-2$ . The roots are $-2\pm5i$ . Their sum is $-4$ and product is $29$ , so $y''+4y'+29y=0$ .	6
(6 marks)		6
Notes
For roots $p\pm qi$ , the trigonometric factor has period $2\pi/\|q\|$ , so the stated period gives $\|q\|=5$ . At $x=3$ , $e^{px}=e^{3p}=e^{-6}$ , hence $p=-2$ . The roots are therefore $-2\pm5i$ . Their sum is $-4$ and product is $4+25=29$ , giving the monic equation $y''+4y'+29y=0$ .
3	The auxiliary equation is $m^2+7m+c=0$ . Its discriminant is $49-4c$ . Two distinct real roots require $c<49/4$ . Their sum is $-7<0$ . Their product is $c$ , so two real roots with negative sum are both negative exactly when $c>0$ . Combining the conditions gives $0<c<49/4$ .	6
(6 marks)		6
Notes
Distinct real roots require $49-4c>0$ , so $c<49/4$ . By Vieta, their sum is always $-7$ , while their product is $c$ . If $c>0$ , the real roots have the same sign and their negative sum makes both negative; if $c\leq0$ , they cannot both be negative. Hence $0<c<49/4$ .

Tier 3 · Hard

Mark scheme for CP-9.6 Tier 3 · Hard
Question	Scheme	Marks
1	Distinct real roots for $p<-7/2$ , a repeated root for $p=-7/2$ , and complex conjugate roots for $p>-7/2$ . At $p=-7/2$ , $y=(A+Bx)e^{5x/2}$ .	6
(6 marks)		6
Notes
The discriminant is $[2(p+1)]^2-4(p^2+4p+8)=-4(2p+7)$ . It is positive, zero or negative according as $p<-7/2$ , $p=-7/2$ or $p>-7/2$ . At equality the repeated root is $-[2(p+1)]/2=-(p+1)=5/2$ , giving $(A+Bx)e^{5x/2}$ .
2	For both roots to be negative, their sum $-a$ must be negative, so $a>0$ . Their product $2a-3$ must also be positive, so $a>3/2$ . Non-oscillation requires $\Delta=a^2-4(2a-3)=(a-2)(a-6)\geq0$ . Thus $a\leq2$ or $a\geq6$ . Combining the conditions gives $\dfrac32<a\leq2$ or $a\geq6$ . At $a=2$ and $a=6$ , the auxiliary roots are repeated. Within the open parts of the stated ranges, the roots are distinct and real.	7
(7 marks)		7
Notes
For both roots to be negative, their sum $-a$ must be negative and product $2a-3$ positive, giving $a>3/2$ . Non-oscillation requires discriminant $a^2-4(2a-3)=(a-2)(a-6)\geq0$ , so $a\leq2$ or $a\geq6$ . Combining conditions gives $(3/2,2]\cup[6,\infty)$ . At $a=2$ and $a=6$ the discriminant is zero; in the open parts it is positive.
3	The auxiliary equation is $m^2+2am+(a+5)=0$ . Its roots are $-a\pm i\sqrt{a+5-a^2}$ when $a+5-a^2>0$ . The angular frequency is $\sqrt{a+5-a^2}$ . Thus the period is $2\pi/\sqrt{a+5-a^2}$ . A period less than $\pi$ requires $\sqrt{a+5-a^2}>2$ . Squaring gives $a^2-a-1<0$ . Therefore $\dfrac{1-\sqrt5}{2}<a<\dfrac{1+\sqrt5}{2}$ .	7
(7 marks)		7
Notes
Completing the quadratic formula gives roots $-a\pm i\sqrt{a+5-a^2}$ in the oscillatory case, so the imaginary part is the angular frequency. The period condition $2\pi/\sqrt{a+5-a^2}<\pi$ is equivalent to $a+5-a^2>4$ , or $a^2-a-1<0$ . This upward-opening quadratic is negative strictly between its roots $(1\pm\sqrt5)/2$ .
4	The auxiliary discriminant is $-8k$ . Oscillation therefore requires $k>0$ . For $k>0$ , the roots are $1-k\pm i\sqrt{2k}$ . The envelope grows when the real part is positive. Thus growth requires $k<1$ . The required set is $0<k<1$ . At $k=0$ there is a repeated root $1$ ; at $k=1$ the roots are $\pm i\sqrt2$ , giving constant-amplitude oscillation.	7
(7 marks)		7
Notes
A negative discriminant gives oscillation, while the real part of the conjugate roots controls the envelope. Intersect the strict conditions $k>0$ and $1-k>0$ , then inspect the two excluded endpoints directly.

CP-9.7 · Solve the equation for simple harmonic motion x'' = -omega^2 x and relate the solution to the motion.

Tier 1 · Easy

Mark scheme for CP-9.7 Tier 1 · Easy
Question	Scheme	Marks
1	$x=A\cos3t+B\sin3t$ ; period $2\pi/3$ .	3
(3 marks)		3
Notes
Comparison with $x''=-\omega^2x$ gives $\omega=3$ . Substitute this angular frequency into the general SHM solution and use $T=2\pi/\omega$ .
2	$\omega=5$ . At equilibrium the speed is maximal, so $15=\omega R=5R$ and the amplitude is $R=3$ metres. The maximum acceleration magnitude is $\omega^2R=25(3)=75$ metres per second squared.	3
(3 marks)		3
Notes
Comparison with $x''=-\omega^2x$ gives $\omega=5$ . At equilibrium the speed has maximum magnitude $\omega R$ , so $15=5R$ and $R=3$ metres. The maximum acceleration magnitude is $\omega^2R=25(3)=75$ metres per second squared.

Tier 2 · Standard

Mark scheme for CP-9.7 Tier 2 · Standard
Question	Scheme	Marks
1	$x=4\cos2t+3\sin2t$ Amplitude $5$ , period $\pi$ , maximum speed $10$ .	6
(6 marks)		6
Notes
Here $\omega=2$ , so $x=A\cos2t+B\sin2t$ . The conditions give $A=4$ and $2B=6$ , hence $B=3$ . The amplitude is $\sqrt{4^2+3^2}=5$ , the period is $2\pi/2=\pi$ , and the maximum speed is $\omega R=2(5)=10$ .
2	Starting at equilibrium in the positive direction with amplitude $5$ gives $x=5\sin(\omega t)$ . The first maximum is a quarter cycle later, so $\omega(\pi/12)=\pi/2$ . Hence $\omega=6$ . $x(t)=5\sin6t$ . The differential equation is $x''=-\omega^2x=-36x$ . The maximum acceleration magnitude is $\omega^2R=36(5)=180$ metres per second squared.	6
(6 marks)		6
Notes
Starting at equilibrium in the positive direction gives $x=5\sin(\omega t)$ . The first maximum occurs after a quarter period, so $\omega(\pi/12)=\pi/2$ and $\omega=6$ . Therefore $x=5\sin6t$ and $x''=-36x$ . The greatest acceleration magnitude is $\omega^2R=36(5)=180$ .
3	The angular frequency is $\omega=2\pi/(2\pi/5)=5$ . Write $x=4\cos(5t+\phi)$ . $x(0)=2$ gives $\cos\phi=1/2$ . Positive initial velocity requires $\sin\phi<0$ , so take $\phi=-\pi/3$ . $x=4\cos(5t-\pi/3)$ and $x''=-25x$ . $x'(0)=-20\sin(-\pi/3)=10\sqrt3$ metres per second. The phase first reaches a value with cosine $-1/2$ at $5t-\pi/3=2\pi/3$ , so $t=\pi/5$ seconds.	7
(7 marks)		7
Notes
The period gives angular frequency $5$ . In amplitude-phase form, the initial displacement requires $\cos\phi=1/2$ . Since $x'(0)=-20\sin\phi$ is positive, choose $\phi=-\pi/3$ . Differentiation gives the SHM equation and initial velocity. As the phase increases from $-\pi/3$ , its first value with cosine $-1/2$ is $2\pi/3$ , a phase increase of $\pi$ , so the first time is $\pi/5$ .

Tier 3 · Hard

Mark scheme for CP-9.7 Tier 3 · Hard
Question	Scheme	Marks
1	$x=\cos(4t+\pi/6)$ $t=\pi/12$ $x''=8$ when $x=-1/2$ .	8
(8 marks)		8
Notes
In $x=A\cos4t+B\sin4t$ , the conditions give $A=\sqrt3/2$ and $4B=-2$ , so $B=-1/2$ . Thus $R=1$ and $x=\cos(4t+\pi/6)$ . The first negative-direction equilibrium crossing has phase $\pi/2$ , so $4t+\pi/6=\pi/2$ and $t=\pi/12$ . Finally $x''=-16x=-16(-1/2)=8$ .
2	The amplitude is $6$ metres. Over one complete oscillation the phase $4t-\pi/3$ increases by $2\pi$ . The period is $2\pi/4=\pi/2$ seconds. $v=dx/dt=24\cos(4t-\pi/3)$ . $v(0)=24\cos(-\pi/3)=12$ metres per second. $x>3$ is equivalent to $\sin(4t-\pi/3)>1/2$ . In one phase cycle this holds for $\pi/6<4t-\pi/3<5\pi/6$ , an interval of phase length $2\pi/3$ . The total time is $(2\pi/3)/4=\pi/6$ seconds.	8
(8 marks)		8
Notes
The amplitude is $6$ and the angular frequency is $4$ , so the period is $2\pi/4=\pi/2$ . Differentiation gives $v=24\cos(4t-\pi/3)$ and hence $v(0)=12$ . Also $x>3$ exactly when $\sin(4t-\pi/3)>1/2$ . Over one phase cycle this occupies $\pi/6<4t-\pi/3<5\pi/6$ , whose phase length is $2\pi/3$ ; division by the angular frequency $4$ gives a total time $\pi/6$ seconds.
3	Successive equilibrium crossings are half a period apart, so the half-period is $2$ seconds. The period is $4$ seconds. The angular frequency is $\omega=\pi/2$ . $x=-7\cos\left(\tfrac\pi2(t-2)\right)$ . At $t=1$ , $x'=-(7\pi/2)<0$ , confirming the stated direction. $x''=-\dfrac{\pi^2}{4}x$ . The maximum speed is $7\pi/2$ metres per second. At $t=2$ , $x=-7$ , so $x''=7\pi^2/4$ metres per second squared.	8
(8 marks)		8
Notes
The direction reverses between the two consecutive equilibrium crossings, so their separation is half a period. This gives period $4$ and angular frequency $\pi/2$ . The negative extreme at the midpoint $t=2$ fixes $x=-7\cos((\pi/2)(t-2))$ . Differentiation verifies the crossing direction and gives $x''=-(\pi^2/4)x$ . The standard SHM speed bound is amplitude times angular frequency.
4	Write $x=a\sin(\omega t+\phi)$ , so $v=a\omega\cos(\omega t+\phi)$ . Eliminating the phase gives $v^2=\omega^2(a^2-x^2)$ . At equilibrium, $\omega a=8$ . At $x=2$ , $36=64-4\omega^2$ . Hence $\omega=\sqrt7$ rad s $^{-1}$ . The amplitude is $a=8\sqrt7/7$ metres. The differential equation is $x''=-7x$ . Writing $x=a\sin\theta$ at the stated instant gives $\theta=\arcsin(\sqrt7/4)$ with increasing phase. The next equilibrium occurs at phase $\pi$ , after $[\pi-\arcsin(\sqrt7/4)]/\sqrt7$ seconds.	9
(9 marks)		9
Notes
The SHM speed-displacement relation follows from the sine solution and its derivative; apply it at the two stated positions. This determines $\omega$ and the rationalised amplitude. A sine phase represents the positive velocity at the given displacement; the next zero after passing the positive extreme occurs when that phase reaches $\pi$ .

CP-9.8 · Model damped oscillations using second order differential equations and interpret their solutions.

Tier 1 · Easy

Mark scheme for CP-9.8 Tier 1 · Easy
Question	Scheme	Marks
1	Underdamped; $x=e^{-3t}(A\cos4t+B\sin4t)$ .	4
(4 marks)		4
Notes
The auxiliary equation has roots $-3\pm4i$ . The non-zero imaginary part produces oscillation, while the negative real part gives the decaying envelope $e^{-3t}$ .
2	The auxiliary equation factorises as $(m+2)(m+5)=0$ . The roots are $-2$ and $-5$ . Two distinct negative real roots give overdamped, non-oscillatory motion. Both modes $e^{-2t}$ and $e^{-5t}$ tend to zero, so $x\to0$ .	4
(4 marks)		4
Notes
The auxiliary equation factorises as $(m+2)(m+5)=0$ . Two distinct negative real roots give an overdamped response. Both exponential modes decay, so every solution approaches equilibrium without sustained oscillation.

Tier 2 · Standard

Mark scheme for CP-9.8 Tier 2 · Standard
Question	Scheme	Marks
1	$c=6$ $x=2(1+t)e^{-3t}$ The displacement tends to $0$ without oscillation.	7
(7 marks)		7
Notes
Critical damping requires $c^2-36=0$ , so positive $c$ gives $c=6$ . The repeated root is $-3$ , hence $x=(A+Bt)e^{-3t}$ . From $x(0)=2$ , $A=2$ ; from $x'(0)=B-3A=-4$ , $B=2$ . Both the exponential and polynomial-exponential term tend to $0$ , and the repeated real root means no oscillation.
2	The auxiliary equation is $m^2+4m+13=0$ . Its roots are $-2\pm3i$ . $x=e^{-2t}(A\cos3t+B\sin3t)$ . $x(0)=3$ gives $A=3$ . $x'(0)=-2A+3B=-6$ gives $B=0$ . $x=3e^{-2t}\cos3t$ . The motion is underdamped because the auxiliary roots are complex; it oscillates with decaying envelope $3e^{-2t}$ .	7
(7 marks)		7
Notes
The auxiliary equation $m^2+4m+13=0$ has roots $-2\pm3i$ , so $x=e^{-2t}(A\cos3t+B\sin3t)$ . The condition $x(0)=3$ gives $A=3$ , and $x'(0)=-2A+3B=-6$ gives $B=0$ . Hence $x=3e^{-2t}\cos3t$ . The complex roots produce oscillation while their negative real part produces decay, so the motion is underdamped.
3	The auxiliary equation is $m^2+18m+81=0$ . $x=(A+Bt)e^{-9t}$ . $x(0)=0$ gives $A=0$ . $x'(0)=9$ gives $B=9$ , so $x=9te^{-9t}$ . $x'=9e^{-9t}(1-9t)$ , so the stationary time is $t=1/9$ . The derivative changes from positive to negative, giving greatest displacement $x(1/9)=1/e$ . The displacement approaches equilibrium without oscillating as $t\to\infty$ .	7
(7 marks)		7
Notes
The repeated auxiliary root is $-9$ , so the critically damped form is $(A+Bt)e^{-9t}$ . The initial conditions give $A=0$ and $B=9$ . Differentiation gives $x'=9e^{-9t}(1-9t)$ , whose sign changes from positive to negative at $t=1/9$ . Substitution gives the maximum $1/e$ , and the exponential factor sends the displacement to zero without oscillation.

Tier 3 · Hard

Mark scheme for CP-9.8 Tier 3 · Hard
Question	Scheme	Marks
1	$x=e^{-2t}(-\dfrac12\cos4t-\dfrac38\sin4t)+\dfrac12\cos2t+\dfrac14\sin2t$ The transient term decays, leaving a steady oscillation of angular frequency $2$ and amplitude $\sqrt5/4$ .	10
(10 marks)		10
Notes
The auxiliary roots are $-2\pm4i$ , so $x_{\mathrm c}=e^{-2t}(A\cos4t+B\sin4t)$ . Try $x_{\mathrm p}=a\cos2t+b\sin2t$ . Substitution gives $16a+8b=10$ and $-8a+16b=0$ , hence $a=1/2$ and $b=1/4$ . From $x(0)=0$ , $A=-1/2$ . At $t=0$ , $x'=-2A+4B+2b=0$ , so $B=-3/8$ . The complementary term tends to $0$ , while the PI remains with amplitude $\sqrt{(1/2)^2+(1/4)^2}=\sqrt5/4$ and angular frequency $2$ .
2	The auxiliary roots are $-1$ and $-4$ , so $x=Ae^{-t}+Be^{-4t}$ . The initial conditions give $A+B=1$ and $-A-4B=2$ . $A=2$ and $B=-1$ , so $x=2e^{-t}-e^{-4t}$ . $x'=-2e^{-t}+4e^{-4t}$ . $x'$ is zero when $e^{3t}=2$ , so $t=\dfrac13\ln2$ . $x'$ changes from positive to negative there, so this is the greatest value for $t\geq0$ . Writing $r=2^{-1/3}$ gives $x_{\max}=2r-r/2=\dfrac32\,2^{-1/3}$ . $x=e^{-t}(2-e^{-3t})>0$ for $t\geq0$ , so the displacement never crosses equilibrium. Both exponential terms tend to $0$ , so $x$ approaches equilibrium.	9
(9 marks)		9
Notes
Roots $-1$ and $-4$ give $x=Ae^{-t}+Be^{-4t}$ . The initial data yield $A+B=1$ and $-A-4B=2$ , so $A=2$ , $B=-1$ . Now $x'=-2e^{-t}+4e^{-4t}$ , which is zero when $e^{3t}=2$ . With $r=2^{-1/3}$ , the maximum is $2r-r/2=3r/2$ . Also $x=e^{-t}(2-e^{-3t})>0$ for $t\geq0$ , and both exponential terms tend to $0$ , proving approach to equilibrium without a crossing.
3	The auxiliary equation $m^2+20m+96=0$ has roots $-8$ and $-12$ . $x=Ae^{-8t}+Be^{-12t}$ . The initial conditions give $A+B=-1$ and $-8A-12B=20$ . Solving gives $A=2$ and $B=-3$ . $x=2e^{-8t}-3e^{-12t}$ . $x=0$ first when $e^{4t}=3/2$ , so $t=\tfrac14\ln(3/2)$ . $x'=-16e^{-8t}+36e^{-12t}$ . The greatest displacement occurs at $t=\tfrac12\ln(3/2)$ . Its value is $2(2/3)^4-3(2/3)^6=32/243$ . The displacement then approaches equilibrium from the positive side without further crossings.	10
(10 marks)		10
Notes
The distinct negative roots give $x=Ae^{-8t}+Be^{-12t}$ . The initial equations yield $A=2$ , $B=-3$ . Equilibrium is crossed when $2e^{-8t}=3e^{-12t}$ , giving $e^{4t}=3/2$ . Setting the derivative to zero gives $e^{4t}=9/4$ , hence $t=\tfrac12\ln(3/2)$ . At this time $e^{-8t}=(2/3)^4$ and $e^{-12t}=(2/3)^6$ , giving $32/243$ . Afterwards the slower positive term dominates and both terms tend to zero.
4	The repeated root gives the auxiliary equation $(m+5)^2=0$ . Hence the differential equation is $x''+10x'+25x=0$ . $x=(A+Bt)e^{-5t}$ . $x(0)=3$ gives $A=3$ . $x(1)=0$ gives $A+B=0$ , so $B=-3$ . Thus $x=3(1-t)e^{-5t}$ . $x'=3e^{-5t}(5t-6)$ . The post-crossing stationary time is $t=6/5$ seconds. $x'$ changes from negative to positive, so the displacement there is a minimum of $-3/(5e^6)$ metres. After this minimum the particle remains below equilibrium and approaches it as $t\to\infty$ .	10
(10 marks)		10
Notes
The specified repeated root fixes the differential equation and solution form. The initial displacement and crossing determine both constants. Differentiation locates the only stationary point after the crossing; its sign change and the exponential limit describe the return toward equilibrium from below.
5	The auxiliary roots are $-3\pm3i$ . $x=e^{-3t}(A\cos3t+B\sin3t)$ . $x(0)=0$ and $x'(0)=6$ give $A=0$ and $B=2$ . Thus $x=2e^{-3t}\sin3t$ . $x'=6e^{-3t}(\cos3t-\sin3t)$ . The first maximum occurs at $t=\pi/12$ seconds. Its value is $\sqrt2e^{-\pi/4}$ metres. The next equilibrium crossing is at $t=\pi/3$ seconds. Successive positive maxima are one period $2\pi/3$ apart, so their ratio is $e^{-2\pi}$ . The oscillations continue with exponentially decreasing amplitude and $x\to0$ .	10
(10 marks)		10
Notes
The complex roots give the damped sine-cosine form, and the initial data leave a single sine term. Setting its derivative to zero locates the first maximum at phase $\pi/4$ and gives its exact height. Zeros come from the sine factor, while the envelope decay over one period gives the maxima ratio.

CP-9.9 · Analyse and interpret models with one independent variable and two dependent variables as a pair of coupled first order simultaneous equations and solve them, e.g. predator-prey models.

Tier 1 · Easy

Mark scheme for CP-9.9 Tier 1 · Easy
Question	Scheme	Marks
1	$u''-2u'-3u=0$	3
(3 marks)		3
Notes
Differentiate $u'=2u+v$ to get $u''=2u'+v'$ . Substitute $v'=3u$ , then rearrange to $u''-2u'-3u=0$ .
2	Solving $p'=-2p$ gives $p=Ce^{-2t}$ . $p(0)=3$ gives $p=3e^{-2t}$ . Substitution into $q'=p-q$ gives $q'+q=3e^{-2t}$ .	3
(3 marks)		3
Notes
Solving $p'=-2p$ gives $p=Ce^{-2t}$ . Since $p(0)=3$ , $p=3e^{-2t}$ . Substitution into $q'=p-q$ gives $q'+q=3e^{-2t}$ .

Tier 2 · Standard

Mark scheme for CP-9.9 Tier 2 · Standard
Question	Scheme	Marks
1	$x=2e^{3t}+e^{-t}$ $y=2e^{3t}-e^{-t}$	7
(7 marks)		7
Notes
From the first equation, $y=\tfrac12(x'-x)$ , so $y'=\tfrac12(x''-x')$ . Substituting into $y'=2x+y$ gives $\tfrac12(x''-x')=2x+\tfrac12(x'-x)$ , i.e. $x''-2x'-3x=0$ . The auxiliary equation $m^2-2m-3=0$ has roots $m=3$ and $m=-1$ , so $x=Ae^{3t}+Be^{-t}$ . Then $y=\tfrac12(x'-x)=Ae^{3t}-Be^{-t}$ . The conditions $x(0)=3$ , $y(0)=1$ give $A+B=3$ and $A-B=1$ , so $A=2$ , $B=1$ : $x=2e^{3t}+e^{-t}$ and $y=2e^{3t}-e^{-t}$ .
2	From $x'=-x-5y$ , $y=-(x'+x)/5$ . Differentiation gives $y'=-(x''+x')/5$ . Substitution in $y'=x+3y$ gives $-(x''+x')=2x-3x'$ . Hence $x''-2x'+2x=0$ . The auxiliary roots are $1\pm i$ , so $x=e^t(A\cos t+B\sin t)$ . $x(0)=5$ gives $A=5$ . $x'(0)=-x(0)-5y(0)=-5$ , so $A+B=-5$ and $B=-10$ . $x=e^t(5\cos t-10\sin t)$ and $y=5e^t\sin t$ .	8
(8 marks)		8
Notes
From $x'=-x-5y$ , $y=-(x'+x)/5$ , so $y'=-(x''+x')/5$ . Substitution in $y'=x+3y$ gives $x''-2x'+2x=0$ . Its roots are $1\pm i$ , hence $x=e^t(A\cos t+B\sin t)$ . The initial conditions give $A=5$ and $x'(0)=-5$ , so $A+B=-5$ and $B=-10$ . Therefore $x=e^t(5\cos t-10\sin t)$ , and $y=-(x'+x)/5=5e^t\sin t$ . Substitution verifies both original equations and both initial conditions.
3	From the first equation, $y=(x'-x)/5$ . Differentiation and substitution give $x''+5x'-66x=0$ . The auxiliary roots are $6$ and $-11$ . $x=Ae^{6t}+Be^{-11t}$ . $x(0)=2$ and $x'(0)=7$ , so $A+B=2$ and $6A-11B=7$ . $A=29/17$ and $B=5/17$ , so $x=(29e^{6t}+5e^{-11t})/17$ . $y=(x'-x)/5=(29e^{6t}-12e^{-11t})/17$ . The $e^{6t}$ terms dominate, so $y/x\to1$ .	8
(8 marks)		8
Notes
Eliminate $y$ using $y=(x'-x)/5$ . Differentiating the first equation and substituting the second gives $x''=-5x'+66x$ . The auxiliary roots are $6$ and $-11$ . The initial system gives $x'(0)=2+5=7$ , from which $A=29/17$ and $B=5/17$ . Recovering $y$ from the elimination formula gives its stated expression. Both leading terms have coefficient $29/17$ , so the ratio tends to $1$ .

Tier 3 · Hard

Mark scheme for CP-9.9 Tier 3 · Hard
Question	Scheme	Marks
1	$A=3e^{-t}+e^{-4t}$ $B=3e^{-t}-2e^{-4t}$ $B$ is greatest at $t=\dfrac13\ln(8/3)$ , with maximum $\dfrac94(3/8)^{1/3}$ . Both amounts tend to $0$ as $t\to\infty$ .	10
(10 marks)		10
Notes
From $A'=-2A+B$ , $B=A'+2A$ . Differentiate and use the second equation to obtain $A''+5A'+4A=0$ . Thus $A=Ce^{-t}+De^{-4t}$ . Now $A(0)=4$ and $A'(0)=-2(4)+1=-7$ , giving $C=3$ , $D=1$ . Hence $B=A'+2A=3e^{-t}-2e^{-4t}$ . For a maximum, $B'=-3e^{-t}+8e^{-4t}=0$ , so $e^{3t}=8/3$ and $t=(1/3)\ln(8/3)$ . Writing $r=e^{-t}=(3/8)^{1/3}$ gives $e^{-4t}=(3/8)r$ , hence $B_{\max}=3r-2(3/8)r=(9/4)r$ . Both exponential terms tend to $0$ in the long term.
2	From $x'=3x-2y+4$ , $y=(3x+4-x')/2$ . Differentiation gives $y'=(3x'-x'')/2$ . Using $y'=2x-2y$ gives $(3x'-x'')/2=2x-(3x+4-x')$ . Rearrangement gives $x''-x'-2x=8$ . The complementary roots are $2$ and $-1$ . A constant particular integral is $x_{\mathrm p}=-4$ . $x=Ae^{2t}+Be^{-t}-4$ . $x(0)=1$ gives $A+B=5$ . The first original equation gives $x'(0)=3(1)-2(3)+4=1$ , so $2A-B=1$ . $A=2$ and $B=3$ , so $x=2e^{2t}+3e^{-t}-4$ . $y=(3x+4-x')/2=e^{2t}+6e^{-t}-4$ .	11
(11 marks)		11
Notes
From the first equation, $y=(3x+4-x')/2$ , so $y'=(3x'-x'')/2$ . Substitution in $y'=2x-2y$ gives $x''-x'-2x=8$ . The complementary roots are $2$ and $-1$ , and a constant PI is $-4$ , so $x=Ae^{2t}+Be^{-t}-4$ . Now $x(0)=1$ gives $A+B=5$ , while the original system gives $x'(0)=1$ and hence $2A-B=1$ . Thus $A=2$ , $B=3$ , and $x=2e^{2t}+3e^{-t}-4$ . Recovering the other variable gives $y=(3x+4-x')/2=e^{2t}+6e^{-t}-4$ ; direct substitution verifies both equations and initial conditions.
3	The equilibrium is $(x,y)=(4,3)$ ; set $X=x-4$ and $Y=y-3$ . Then $X'=-8X+Y$ and $Y'=2X-9Y$ . Eliminating $Y$ gives $X''+17X'+70X=0$ . The auxiliary roots are $-7$ and $-10$ . $X=Ae^{-7t}+Be^{-10t}$ . $X(0)=-3$ and $X'(0)=24$ give $A=-2$ and $B=-1$ . $x=4-2e^{-7t}-e^{-10t}$ . $y=3-2e^{-7t}+2e^{-10t}$ . $y'=14e^{-7t}-20e^{-10t}$ , so the minimum occurs at $t=\tfrac13\ln(10/7)$ and equals $3-\tfrac35(7/10)^{7/3}$ . $x$ increases from $1$ , while the displayed minimum of $y$ is greater than $12/5$ , so both quantities stay positive. As $t\to\infty$ , $(x,y)\to(4,3)$ .	11
(11 marks)		11
Notes
Subtract the equilibrium to make the system homogeneous. Since $Y=X'+8X$ , differentiation and substitution give $X''+17X'+70X=0$ , with roots $-7$ and $-10$ . The shifted initial values are $X(0)=-3$ , $Y(0)=0$ , hence $X'(0)=24$ , giving $A=-2$ , $B=-1$ . Recover $Y=X'+8X$ to find $y$ . Its derivative changes from negative to positive when $e^{3t}=10/7$ . At that time $e^{-7t}=(7/10)^{7/3}$ and $e^{-10t}=(7/10)^{10/3}$ , giving the stated minimum. Also $x'=14e^{-7t}+10e^{-10t}>0$ , and the minimum bound for $y$ is positive. Both exponentials disappear in the limit.
4	From the first equation, $y=(x'+2x)/2$ . Differentiation and use of $y'=4x$ give $x''+2x'-8x=0$ . The auxiliary roots are $2$ and $-4$ . $x=Ae^{2t}+Be^{-4t}$ . $y=2Ae^{2t}-Be^{-4t}$ . The initial conditions give $A=1$ and $B=2$ . Thus $x=e^{2t}+2e^{-4t}$ and $y=2e^{2t}-2e^{-4t}$ . $x'=2e^{2t}-8e^{-4t}$ , so the stationary time is $t=\tfrac13\ln2$ . $x'$ changes from negative to positive, so this is the least value. The least value is $3/\sqrt[3]{2}$ , and $\lim_{t\to\infty}y/x=2$ .	10
(10 marks)		10
Notes
Eliminate $y$ to obtain a second-order equation for $x$ , then recover $y$ from the first equation. The initial data determine the two exponential coefficients. Differentiating $x$ gives its unique minimum, where $e^{6t}=4$ , and the growing exponential terms determine the limiting ratio.
5	$(3x+y)'=3(-3x+y)+(9x-3y)=0$ . Thus $3x+y=3x(0)+y(0)=11$ . Let $v=y-3x$ . $v'=y'-3x'=-6v$ . $v(0)=2-9=-7$ . Hence $v=-7e^{-6t}$ . Solving $3x+y=11$ and $y-3x=-7e^{-6t}$ gives $x=11/6+(7/6)e^{-6t}$ . The corresponding solution is $y=11/2-(7/2)e^{-6t}$ . $x=y$ gives $e^{-6t}=11/14$ . Therefore $t=\tfrac16\ln(14/11)$ . As $t\to\infty$ , $(x,y)\to(11/6,11/2)$ .	11
(11 marks)		11
Notes
The conserved combination and a decaying difference reduce the system to two elementary equations. Solving the resulting simultaneous algebraic equations gives $x$ and $y$ , after which the equality time and limiting state follow.