What is 8.7: Separable differential equations?

For dy/ dx=g(x)h(y), rearrange to place all y-terms with dy and all x-terms with dx, then integrate both sides.

How many marks are 8.7 questions worth?

8.7 practice drills on this page are worth 3–9 marks each; marks in a real exam vary by question.

Which past exam papers has 8.7 appeared in?

We have not yet indexed a verified past-paper appearance for 8.7 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

8.7

Evaluate the analytical solution of simple first order differential equations with separable variables, including finding particular solutions.

Draft — not yet indexed

Separable differential equations

Worked answers and methods for 8.7 on Edexcel A-level Maths 9MA0.

Explanation

For $\frac{\mathrm dy}{\mathrm dx}=g(x)h(y)$, rearrange to place all $y$ -terms with $\mathrm dy$ and all $x$ -terms with $\mathrm dx$ , then integrate both sides.
Include a constant after integration and use the initial condition to determine it.
A family of solutions can be sketched from its equilibrium curves, initial values, gradients and long-term behaviour.
When logarithms arise, exponentiate carefully and use the stated domain or initial condition to choose any required sign or branch.
A common error is to divide by a factor involving $y$ without checking whether doing so loses a constant equilibrium solution.

Worked example

Solve $\frac{\mathrm dy}{\mathrm dx}=\frac{x+1}{y}$ , given that $y=2$ when $x=0$ and $y>0$ .

1.Separate: $y\,\mathrm dy=(x+1)\,\mathrm dx$ .
2.Integration gives $\frac12y^2=\frac12x^2+x+C$ , so $y^2=x^2+2x+C_1$ .
3.The condition $y(0)=2$ gives $C_1=4$ .
4.Since $y>0$ , $y=\sqrt{x^2+2x+4}$ .

Answer: $y=\sqrt{x^2+2x+4}$

Common mistakes

Don't rearrange $\mathrm dy/\mathrm dx=f(x)/g(y)$ as $f(x)\,\mathrm dy=g(y)\,\mathrm dx$ .
Don't separate variables but lose the integration constant or select a branch inconsistent with the initial condition.

Exam tip

After separation and integration, use the initial condition before choosing the branch required by the stated domain.

Worked practice

Tier 1 · Easy

Solve

\frac{\mathrm dy}{\mathrm dx}=2xy

, given that

y=3

when

x=0

(4)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$y=3e^{x^2}$	4
Notes
Separate variables: $\frac1y\,\mathrm dy=2x\,\mathrm dx$ . Integrating gives $\ln\|y\|=x^2+C$ , so $y=Ae^{x^2}$ . The condition $y(0)=3$ gives $A=3$ , hence $y=3e^{x^2}$ .

(4 marks)

Tier 2 · Standard

For

\dfrac{\mathrm dy}{\mathrm dx}=3x(y-2)

, find the general non-equilibrium solution and state the equilibrium solution excluded when the variables are separated.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$y=2+Ae^{3x^2/2}$ , where $A\ne0$ Equilibrium solution: $y=2$	4
Notes
Before dividing by $y-2$ , note that $y=2$ is a constant equilibrium solution. For $y\ne2$ , separate to obtain $\mathrm dy/(y-2)=3x\,\mathrm dx$ . Integration gives $\ln\|y-2\|=3x^2/2+C$ , so $y-2=Ae^{3x^2/2}$ . Hence the non-equilibrium family is $y=2+Ae^{3x^2/2}$ with $A\ne0$ , alongside the excluded equilibrium solution $y=2$ .

(4 marks)

Tier 3 · Hard

Solve

\frac{\mathrm dy}{\mathrm dx}=y(4-y)

, given that

y=1

when

x=0

. State the equilibrium solutions and describe the particular solution's long-term behaviour.

(9)

(Total for Question 3 is 9 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$y=\frac4{1+3e^{-4x}}$ Equilibrium solutions $y=0$ and $y=4$ ; the particular solution increases towards $y=4$	9
Notes
The constant solutions lost on division are $y=0$ and $y=4$ . Otherwise separate and integrate to obtain $\frac14(\ln\|y\|-\ln\|4-y\|)=x+C$ , so $\frac{y}{4-y}=Ae^{4x}$ . The initial condition gives $A=1/3$ , hence $y=\frac4{1+3e^{-4x}}$ . It starts at $1$ , has positive gradient while $0<y<4$ , and tends to the upper equilibrium $4$ .

(9 marks)

Tier 1 · Easy

Solve

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2}{2y}

, given that

y=2

when

x=0

and that

y>0

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$y=\sqrt{x^3+4}$	3
Notes
Separate variables: $2y\,\mathrm dy=3x^2\,\mathrm dx$ . Integration gives $y^2=x^3+C$ . The initial condition gives $C=4$ , and the condition $y>0$ selects $y=\sqrt{x^3+4}$ .

(3 marks)

Tier 2 · Standard

Solve

\dfrac{\mathrm dy}{\mathrm dx}=(2x+1)e^{-y}

, given that

y=\ln2

when

x=0

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$y=\ln(x^2+x+2)$	4
Notes
Separate variables to obtain $e^y\,\mathrm dy=(2x+1)\,\mathrm dx$ . Integration gives $e^y=x^2+x+C$ . Since $y=\ln2$ at $x=0$ , $C=2$ . Taking logarithms gives $y=\ln(x^2+x+2)$ .

(4 marks)

Tier 3 · Hard

Solve

\dfrac{\mathrm dy}{\mathrm dx}=x(y-1)^2

, given that

y=0

when

x=0

. State the equilibrium solution and the limiting value of the particular solution as

x\to\infty

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$y=\dfrac{x^2}{x^2+2}$ Equilibrium solution $y=1$ ; the particular solution tends to $1$	6
Notes
The equilibrium solution lost by division is $y=1$ . Otherwise $(y-1)^{-2}\,\mathrm dy=x\,\mathrm dx$ , giving $-1/(y-1)=x^2/2+C$ . The initial condition gives $C=1$ , so $1/(1-y)=1+x^2/2$ and hence $y=x^2/(x^2+2)$ . As $x\to\infty$ , this tends to $1$ from below.

(6 marks)

Tier 2 · Standard

Solve

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+1}{y^2}

, given that

y=1

when

x=0

and that

y>0

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$y=(3x^2+3x+1)^{1/3}$	4
Notes
Separate variables to obtain $y^2\,\mathrm dy=(2x+1)\,\mathrm dx$ . Integration gives $y^3/3=x^2+x+C$ . The initial condition gives $C=1/3$ , so $y^3=3x^2+3x+1$ . The stated positive branch is $y=(3x^2+3x+1)^{1/3}$ .

(4 marks)

Tier 3 · Hard

Solve

\dfrac{\mathrm dy}{\mathrm dx}=(y+2)\sin x

, given that

y=-1

when

x=0

. State the equilibrium solution. Hence find all values of

x

0\leq x\leq2\pi

for which

y=e-2

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$y=e^{1-\cos x}-2$ Equilibrium solution $y=-2$ $x=\dfrac{\pi}{2},\dfrac{3\pi}{2}$	6
Notes
The equilibrium solution lost on division is $y=-2$ . Otherwise $\mathrm dy/(y+2)=\sin x\,\mathrm dx$ , so $\ln\|y+2\|=-\cos x+C$ . The initial condition gives $C=1$ and the positive sign, hence $y=e^{1-\cos x}-2$ . Setting this equal to $e-2$ gives $e^{1-\cos x}=e$ , so $\cos x=0$ . The two solutions in the stated interval are $\pi/2$ and $3\pi/2$ .

(6 marks)

Tier 3 · Hard

Solve

\dfrac{\mathrm dy}{\mathrm dx}=(1+y^2)\cos x

, given that

y=0

when

x=0

. For

0\leq x\leq2\pi

, find the maximum and minimum values of

y

and the values of

x

at which they occur.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$y=\tan(\sin x)$ Maximum $y=\tan1$ at $x=\dfrac{\pi}{2}$ ; minimum $y=-\tan1$ at $x=\dfrac{3\pi}{2}$	6
Notes
Separating gives $\mathrm dy/(1+y^2)=\cos x\,\mathrm dx$ , so $\arctan y=\sin x+C$ . The initial condition gives $C=0$ . Since $-1\leq\sin x\leq1$ lies wholly inside the principal range $(-\pi/2,\pi/2)$ of $\arctan$ , inversion gives the single continuous branch $y=\tan(\sin x)$ . The tangent function is increasing on this range, so $y$ is largest when $\sin x=1$ , at $x=\pi/2$ , and smallest when $\sin x=-1$ , at $x=3\pi/2$ .

(6 marks)

Q10

Tier 3 · Hard

10.

Solve

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^x}{y^2+1}

, given that

y=0

when

x=0

, leaving your solution in implicit form. Hence find the value of

x

for which

y=2

, and explain why this value is unique.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\dfrac{y^3}{3}+y=e^x-1$ $x=\ln\left(\dfrac{17}{3}\right)$ The value is unique because $\mathrm dy/\mathrm dx>0$ for all real $x$	6
Notes
Separating gives $(y^2+1)\,\mathrm dy=e^x\,\mathrm dx$ . Integration gives $y^3/3+y=e^x+C$ , and the initial condition gives $C=-1$ . When $y=2$ , $y^3/3+y=14/3$ , so $e^x=17/3$ and $x=\ln(17/3)$ . Since both $e^x$ and $y^2+1$ are positive, $\mathrm dy/\mathrm dx>0$ for every real $x$ ; the solution is strictly increasing and can take the value $2$ only once.

(6 marks)

Verified exam appearances

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