Further differential equations (Further Pure 1)

FP1-3 Further differential equations (Further Pure 1) — revision question pack

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

Checked against Edexcel 9FM0 section FP1-3. 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.

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FP1-3.1 · Use of Taylor series method for series solution of differential equations.

Explanation

When a differential equation cannot be solved in closed form, a Taylor series about the point where the initial conditions are given still produces a usable local solution. The method is mechanical:
substitute the initial values into the equation to get the first unknown derivative, differentiate the whole equation with respect to $x$ , substitute again, and repeat until you have every derivative you need.
Assemble them with $y=y(a)+y'(a)(x-a)+\dfrac{y''(a)}{2!}(x-a)^2+\cdots$ . For a first-order equation $y'=f(x,y)$ you differentiate $f$ using the chain rule, remembering that $y$ is a function of $x$ , so $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(y^2\right)=2yy'$ .
For a second-order equation both $y(a)$ and $y'(a)$ are given and the equation supplies $y''(a)$ directly.
Keep a running table of the values at $x=a$ ; that table, not the algebra, is what you assemble the series from.

Worked example

Find the series solution in ascending powers of $x$ , as far as the term in $x^4$ , of $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+x\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=0$ , given that $y=0$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}=1$ at $x=0$ .

1.Rearranged, $y''=-xy'-y$ , so at $x=0$ , $y''=0-0=0$ .
2.Differentiating, $y'''=-2y'-xy''$ , so $y'''(0)=-2$ .
3.Differentiating again, $y^{(4)}=-3y''-xy'''$ , so $y^{(4)}(0)=0$ .
4.Assembling: $y=0+x+0-\dfrac{2x^3}{6}+0=x-\dfrac{x^3}{3}$ .

Answer: $y=x-\dfrac{x^3}{3}+\cdots$ , with the $x^2$ and $x^4$ terms both zero.

Common mistakes

Don't fall into the trap of differentiating $y^2$ as $2y$ instead of $2yy'$ .
Don't fall into the trap of substituting $x=0$ before differentiating the whole equation, which destroys the terms needed next.
Don't fall into the trap of dividing the derivative values by $n$ rather than $n!$ when assembling the series.

Exam tip

Never substitute the initial values into an expression you still need to differentiate; keep one general line and one numerical line side by side.

Tier 1 · Easy

1.
Given that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y$ and $y=1$ at $x=0$ , find the series solution for $y$ in ascending powers of $x$ as far as the term in $x^3$ .
(4)
(Total for Question 1 is 4 marks)
2.
Given that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=y^2$ and $y=1$ at $x=0$ , find the series solution for $y$ as far as the term in $x^3$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Given that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x^2+y^2$ and $y=1$ at $x=0$ , find the series solution for $y$ as far as the term in $x^4$ .
(6)
(Total for Question 1 is 6 marks)
2.
Given that $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\mathrm{d}y}{\mathrm{d}x}+2y$ with $y=1$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ at $x=0$ , find the series solution as far as the term in $x^4$ .
(6)
(Total for Question 2 is 6 marks)
3.
Given that $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=x\dfrac{\mathrm{d}y}{\mathrm{d}x}+y$ with $y=1$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}=1$ at $x=0$ , find the series solution as far as the term in $x^5$ .
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Given that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\sin(x+y)$ and $y=0$ at $x=0$ , find the series solution for $y$ as far as the term in $x^3$ .
(6)
(Total for Question 1 is 6 marks)
2.
Given that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y^2$ and $y=0$ at $x=0$ , find the series solution for $y$ as far as the term in $x^5$ .
(6)
(Total for Question 2 is 6 marks)
3.
Given that $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=y^2+x$ with $y=1$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ at $x=0$ , find the series solution as far as the term in $x^4$ .
(6)
(Total for Question 3 is 6 marks)
4.
Given that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x-y^2$ and $y=2$ at $x=1$ , find the series solution for $y$ in ascending powers of $(x-1)$ as far as the term in $(x-1)^3$ .
(6)
(Total for Question 4 is 6 marks)
5.
Using the series solution $y=2-3(x-1)+\dfrac{13}{2}(x-1)^2-\dfrac{35}{3}(x-1)^3$ of $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x-y^2$ with $y(1)=2$ , estimate $y(1.1)$ to $4$ decimal places and comment on the accuracy of the estimate.
(5)
(Total for Question 5 is 5 marks)

FP1-3.2 · Differential equations reducible by means of a given substitution.

Explanation

The question always supplies the substitution; your job is to carry it through correctly and land on one of the standard forms from Core Pure section 9 — a first-order linear equation solved by an integrating factor, a separable equation, or a second-order linear equation with constant coefficients. Three patterns dominate.
First, $z=y^{1-n}$ turns $y'+P(x)y=Q(x)y^{n}$ into a linear equation in $z$ ; the working starts by dividing through by $y^{n}$ .
Second, $y=vx$ turns a homogeneous equation into a separable one in $v$ and $x$ , using $\dfrac{\mathrm{d}y}{\mathrm{d}x}=v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}$ .
Third, $x=\mathrm{e}^{u}$ turns an Euler equation into constant coefficients, using $x\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y}{\mathrm{d}u}$ and $x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}-\dfrac{\mathrm{d}y}{\mathrm{d}u}$ .
Always convert the final answer back to the original variables, and apply any boundary condition only after that conversion.

Worked example

Use the substitution $z=y^{2}$ to solve $2y\dfrac{\mathrm{d}y}{\mathrm{d}x}+y^{2}=x$ .

1. $z=y^2$ gives $\dfrac{\mathrm{d}z}{\mathrm{d}x}=2y\dfrac{\mathrm{d}y}{\mathrm{d}x}$ , so the equation becomes $\dfrac{\mathrm{d}z}{\mathrm{d}x}+z=x$ .
2.The integrating factor is $\mathrm{e}^{x}$ , so $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(z\mathrm{e}^{x}\right)=x\mathrm{e}^{x}$ .
3.Integrating by parts, $z\mathrm{e}^{x}=(x-1)\mathrm{e}^{x}+A$ , so $z=x-1+A\mathrm{e}^{-x}$ .

Answer: $y^{2}=x-1+A\mathrm{e}^{-x}$ .

Common mistakes

Don't fall into the trap of failing to divide through by $y^{n}$ before substituting in a Bernoulli equation, so the new variable never appears cleanly.
Don't fall into the trap of using $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x\dfrac{\mathrm{d}v}{\mathrm{d}x}$ instead of $v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}$ for $y=vx$ .
Don't fall into the trap of leaving the answer in terms of the substituted variable instead of returning to $y$ and $x$ .

Exam tip

Write the derivative of the substitution on its own line before touching the equation; almost every lost mark in this topic comes from that one differentiation.

Tier 1 · Easy

1.
Use the substitution $z=y^{-1}$ to show that $\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=xy^{2}$ reduces to $\dfrac{\mathrm{d}z}{\mathrm{d}x}-z=-x$ .
(3)
(Total for Question 1 is 3 marks)
2.
Use the substitution $y=vx$ to show that $x\dfrac{\mathrm{d}y}{\mathrm{d}x}=y+x\tan\dfrac{y}{x}$ reduces to $x\dfrac{\mathrm{d}v}{\mathrm{d}x}=\tan v$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Use the substitution $z=y^{-1}$ to find the general solution of $\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=xy^{2}$ .
(6)
(Total for Question 1 is 6 marks)
2.
Use the substitution $x=\mathrm{e}^{u}$ to find the general solution of $x^{2}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+x\dfrac{\mathrm{d}y}{\mathrm{d}x}-4y=0$ for $x>0$ .
(6)
(Total for Question 2 is 6 marks)
3.
Use the substitution $u=\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to find the general solution of $x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+\dfrac{\mathrm{d}y}{\mathrm{d}x}=x$ for $x>0$ .
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Use the substitution $y=vx$ to solve $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{x^{2}+y^{2}}{xy}$ given that $y=2$ when $x=1$ , and express $y$ in terms of $x$ .
(7)
(Total for Question 1 is 7 marks)
2.
Use the substitution $z=\sin y$ to find the general solution of $\cos y\dfrac{\mathrm{d}y}{\mathrm{d}x}+\dfrac{\sin y}{x}=x$ for $x>0$ .
(7)
(Total for Question 2 is 7 marks)
3.
Use the substitution $u=xy$ to find the general solution of $x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+2\dfrac{\mathrm{d}y}{\mathrm{d}x}+xy=0$ for $x>0$ .
(7)
(Total for Question 3 is 7 marks)
4.
Use the substitution $u=\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to solve $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=1+\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}$ given that $y=0$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ at $x=0$ .
(7)
(Total for Question 4 is 7 marks)
5.
Use the substitution $x=\mathrm{e}^{u}$ to find the general solution of $x^{2}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-3x\dfrac{\mathrm{d}y}{\mathrm{d}x}+4y=x^{2}$ for $x>0$ .
(8)
(Total for Question 5 is 8 marks)

Answer key

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

FP1-3.1 · Use of Taylor series method for series solution of differential equations.

Tier 1 · Easy

Mark scheme for FP1-3.1 Tier 1 · Easy
Question	Scheme	Marks
1	$y'(0)=0+1=1$ $y''=1+y'$ so $y''(0)=2$ ; $y'''=y''$ so $y'''(0)=2$ $y=1+x+x^2+\dfrac{x^3}{3}+\cdots$	4
(4 marks)		4
Notes
Substituting gives $y'(0)=1$ ; differentiating the equation gives $y''=1+y'$ and $y'''=y''$ , so the values at $0$ are $2$ and $2$ , and the coefficients are $\tfrac{2}{2!}=1$ and $\tfrac{2}{3!}=\tfrac13$ . Independent check: the exact solution is $y=2\mathrm{e}^{x}-x-1$ , whose expansion is $2\left(1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}\right)-x-1=1+x+x^2+\tfrac{x^3}{3}$ .
2	$y'(0)=1$ $y''=2yy'$ so $y''(0)=2$ $y'''=2\left(y'\right)^2+2yy''$ so $y'''(0)=2+4=6$ $y=1+x+x^2+x^3+\cdots$	4
(4 marks)		4
Notes
Differentiating with the chain rule gives $y''=2yy'$ and $y'''=2(y')^2+2yy''$ , with values $2$ and $6$ at $x=0$ ; dividing by $2!$ and $3!$ gives coefficients $1$ and $1$ . Independent check: separating variables gives $y=\dfrac1{1-x}$ , whose binomial expansion is $1+x+x^2+x^3+\cdots$ .

Tier 2 · Standard

Mark scheme for FP1-3.1 Tier 2 · Standard
Question	Scheme	Marks
1	$y'(0)=1$ $y''=2x+2yy'$ so $y''(0)=2$ $y'''=2+2\left(y'\right)^2+2yy''$ so $y'''(0)=8$ $y^{(4)}=6y'y''+2yy'''$ so $y^{(4)}(0)=12+16=28$ $y=1+x+x^2+\dfrac{4x^3}{3}+\dfrac{7x^4}{6}+\cdots$	6
(6 marks)		6
Notes
Each differentiation uses the chain rule on $y^2$ , giving the derivative values $1,2,8,28$ at $x=0$ ; dividing by $2!,3!,4!$ gives $1$ , $\tfrac86=\tfrac43$ and $\tfrac{28}{24}=\tfrac76$ . Independent check by fourth-order numerical integration with step $10^{-6}$ : the solution at $x=0.1$ is $1.111463$ , and the series gives $1+0.1+0.01+0.001333+0.000117=1.111450$ , agreeing to the order of the first neglected term.
2	$y''(0)=0+2=2$ $y'''=y''+2y'$ so $y'''(0)=2$ $y^{(4)}=y'''+2y''$ so $y^{(4)}(0)=6$ $y=1+x^2+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$	6
(6 marks)		6
Notes
The equation gives $y''(0)=2$ directly, and each further differentiation reuses the same relation, giving $2$ and $6$ . Independent check: the auxiliary equation $m^2-m-2=0$ has roots $2$ and $-1$ , so $y=\tfrac13\mathrm{e}^{2x}+\tfrac23\mathrm{e}^{-x}$ , whose expansion is $1+x^2+\tfrac{x^3}{3}+\tfrac{x^4}{4}+\cdots$ .
3	$y''(0)=0+1=1$ $y'''=2y'+xy''$ so $y'''(0)=2$ $y^{(4)}=3y''+xy'''$ so $y^{(4)}(0)=3$ $y^{(5)}=4y'''+xy^{(4)}$ so $y^{(5)}(0)=8$ $y=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{8}+\dfrac{x^5}{15}+\cdots$	6
(6 marks)		6
Notes
Differentiating the product $xy'$ each time raises the coefficient by one, giving the pattern $y^{(n+2)}=ny^{(n)}+\cdots$ and the values $1,2,3,8$ . Dividing by the factorials gives $\tfrac12$ , $\tfrac26=\tfrac13$ , $\tfrac3{24}=\tfrac18$ and $\tfrac8{120}=\tfrac1{15}$ . Independent check: the values satisfy the equation term by term, since substituting the series into $y''-xy'-y$ makes every coefficient up to $x^3$ vanish.

Tier 3 · Hard

Mark scheme for FP1-3.1 Tier 3 · Hard
Question	Scheme	Marks
1	$y'(0)=\sin0=0$ $y''=\cos(x+y)\left(1+y'\right)$ so $y''(0)=1$ $y'''=-\sin(x+y)\left(1+y'\right)^2+\cos(x+y)y''$ so $y'''(0)=1$ $y=\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots$	6
(6 marks)		6
Notes
Differentiating $\sin(x+y)$ requires the chain rule with inner derivative $1+y'$ , and the second differentiation needs the product rule as well. Independent check by fourth-order numerical integration with step $10^{-4}$ : at $x=0.2$ the numerical solution is $0.021314$ , and the series gives $0.02+0.001333=0.021333$ .
2	$y'(0)=0$ and $y''=1+2yy'$ so $y''(0)=1$ $y'''=2\left(y'\right)^2+2yy''$ so $y'''(0)=0$ $y^{(4)}=6y'y''+2yy'''$ so $y^{(4)}(0)=0$ $y^{(5)}=6\left(y''\right)^2+8y'y'''+2yy^{(4)}$ so $y^{(5)}(0)=6$ $y=\dfrac{x^2}{2}+\dfrac{x^5}{20}+\cdots$	6
(6 marks)		6
Notes
Because $y(0)=y'(0)=0$ , every term containing $y$ or $y'$ vanishes at the origin until the square of $y''$ appears at the fifth derivative, where $6\left(y''\right)^2=6$ . Dividing by $5!$ gives $\tfrac6{120}=\tfrac1{20}$ . Independent check by substituting $y=\tfrac{x^2}{2}+cx^5$ into the equation: the left side is $x+5cx^4$ and the right side is $x+\tfrac{x^4}{4}$ , so $c=\tfrac1{20}$ .
3	$y''(0)=1+0=1$ $y'''=2yy'+1$ so $y'''(0)=1$ $y^{(4)}=2\left(y'\right)^2+2yy''$ so $y^{(4)}(0)=2$ $y=1+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{12}+\cdots$	6
(6 marks)		6
Notes
The equation gives $y''(0)=1$ ; differentiating gives $y'''=2yy'+1$ and $y^{(4)}=2(y')^2+2yy''$ , with values $1$ and $2$ at the origin. Dividing by $3!$ and $4!$ gives $\tfrac16$ and $\tfrac2{24}=\tfrac1{12}$ . Independent check by substitution: with $y=1+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\tfrac{x^4}{12}$ , the left side is $1+x+x^2+\cdots$ and $y^2+x=1+x+x^2+\cdots$ , matching to the order retained.
4	$y'(1)=1-4=-3$ $y''=1-2yy'$ so $y''(1)=1+12=13$ $y'''=-2\left(y'\right)^2-2yy''$ so $y'''(1)=-18-52=-70$ $y=2-3(x-1)+\dfrac{13}{2}(x-1)^2-\dfrac{35}{3}(x-1)^3+\cdots$	6
(6 marks)		6
Notes
Substituting the initial values gives $y'(1)=-3$ ; two further differentiations give $13$ and $-70$ , and dividing by $2!$ and $3!$ gives $\tfrac{13}{2}$ and $-\tfrac{70}{6}=-\tfrac{35}{3}$ . Independent check: the recurrence values reproduce the equation when the series is substituted, and each coefficient alternates in sign as the $-y^2$ term dominates.
5	With $x-1=0.1$ the terms are $2$ , $-0.3$ , $0.065$ and $-0.0116667$ $y(1.1)\approx1.7533$ A fourth-degree term would be needed for greater accuracy The terms are not yet decreasing quickly, so the estimate is reliable only to about $2$ decimal places	5
(5 marks)		5
Notes
Substituting $x-1=0.1$ gives $2-0.3+0.065-0.0116667=1.7533333$ , that is $1.7533$ to $4$ decimal places. Independent check by fourth-order numerical integration of the same initial-value problem with step $10^{-5}$ : $y(1.1)=1.7551$ , so the truncated series is in error by about $0.0018$ , consistent with the size of the omitted quartic term and confirming that only about two decimal places are trustworthy.

FP1-3.2 · Differential equations reducible by means of a given substitution.

Tier 1 · Easy

Mark scheme for FP1-3.2 Tier 1 · Easy
Question	Scheme	Marks
1	Dividing by $y^2$ gives $y^{-2}\dfrac{\mathrm{d}y}{\mathrm{d}x}+y^{-1}=x$ $z=y^{-1}$ gives $\dfrac{\mathrm{d}z}{\mathrm{d}x}=-y^{-2}\dfrac{\mathrm{d}y}{\mathrm{d}x}$ Substituting gives $-\dfrac{\mathrm{d}z}{\mathrm{d}x}+z=x$ , that is $\dfrac{\mathrm{d}z}{\mathrm{d}x}-z=-x$	3
(3 marks)		3
Notes
Divide the Bernoulli equation by $y^2$ so that the combination $y^{-2}y'$ appears, then replace it by $-z'$ and $y^{-1}$ by $z$ . Independent check: reversing the substitution in $z'-z=-x$ gives $-y^{-2}y'-y^{-1}=-x$ , and multiplying by $-y^2$ returns the original equation.
2	$y=vx$ gives $\dfrac{\mathrm{d}y}{\mathrm{d}x}=v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}$ The equation becomes $x\left(v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}\right)=vx+x\tan v$ Cancelling $vx$ and dividing by $x$ gives $x\dfrac{\mathrm{d}v}{\mathrm{d}x}=\tan v$	3
(3 marks)		3
Notes
The product rule gives $y'=v+xv'$ , and $\tfrac{y}{x}=v$ , so the right side is $vx+x\tan v$ ; the $vx$ terms cancel and one factor of $x$ divides out. Independent check: the reduced equation separates as $\cot v\,\mathrm{d}v=\tfrac{\mathrm{d}x}{x}$ , giving $\ln\|\sin v\|=\ln\|x\|+c$ , so $\sin\tfrac{y}{x}=Ax$ , and differentiating that implicitly returns the original equation.

Tier 2 · Standard

Mark scheme for FP1-3.2 Tier 2 · Standard
Question	Scheme	Marks
1	The equation becomes $\dfrac{\mathrm{d}z}{\mathrm{d}x}-z=-x$ The integrating factor is $\mathrm{e}^{-x}$ $z\mathrm{e}^{-x}=-\displaystyle\int x\mathrm{e}^{-x}\,\mathrm{d}x=(x+1)\mathrm{e}^{-x}+A$ $z=x+1+A\mathrm{e}^{x}$ $y=\dfrac{1}{x+1+A\mathrm{e}^{x}}$	6
(6 marks)		6
Notes
After the Bernoulli reduction the integrating factor $\mathrm{e}^{-x}$ gives $\left(z\mathrm{e}^{-x}\right)'=-x\mathrm{e}^{-x}$ , and integrating by parts gives $(x+1)\mathrm{e}^{-x}+A$ . Independent check by substitution: with $y=\left(x+1+A\mathrm{e}^{x}\right)^{-1}$ , $y'=-\left(1+A\mathrm{e}^{x}\right)y^2$ , so $y'+y-xy^2=y^2\left[-1-A\mathrm{e}^{x}+x+1+A\mathrm{e}^{x}-x\right]=0$ .
2	$x\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y}{\mathrm{d}u}$ and $x^{2}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=\dfrac{\mathrm{d}^{2}y}{\mathrm{d}u^{2}}-\dfrac{\mathrm{d}y}{\mathrm{d}u}$ The equation becomes $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}u^{2}}-4y=0$ The auxiliary equation $m^{2}-4=0$ gives $m=\pm2$ $y=A\mathrm{e}^{2u}+B\mathrm{e}^{-2u}$ $y=Ax^{2}+\dfrac{B}{x^{2}}$	6
(6 marks)		6
Notes
Substituting the two standard Euler identities cancels the first-derivative terms, leaving $y_{uu}-4y=0$ with solution $A\mathrm{e}^{2u}+B\mathrm{e}^{-2u}$ ; since $\mathrm{e}^{u}=x$ this is $Ax^2+Bx^{-2}$ . Independent check by direct substitution: for $y=x^{k}$ the equation gives $k(k-1)+k-4=k^2-4=0$ , so $k=\pm2$ , matching the two basis solutions.
3	The equation becomes $x\dfrac{\mathrm{d}u}{\mathrm{d}x}+u=x$ , that is $\dfrac{\mathrm{d}}{\mathrm{d}x}(xu)=x$ $xu=\dfrac{x^{2}}{2}+A$ , so $u=\dfrac{x}{2}+\dfrac{A}{x}$ Integrating, $y=\dfrac{x^{2}}{4}+A\ln x+B$	6
(6 marks)		6
Notes
Reducing the order turns the equation into a first-order linear equation whose left side is already the exact derivative of $xu$ . Integrating twice gives the general solution with two arbitrary constants. Independent check by substitution: for $y=\tfrac{x^2}{4}+A\ln x+B$ , $y'=\tfrac{x}{2}+\tfrac{A}{x}$ and $y''=\tfrac12-\tfrac{A}{x^2}$ , so $xy''+y'=\tfrac{x}{2}-\tfrac{A}{x}+\tfrac{x}{2}+\tfrac{A}{x}=x$ .

Tier 3 · Hard

Mark scheme for FP1-3.2 Tier 3 · Hard
Question	Scheme	Marks
1	$v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}=\dfrac{1+v^{2}}{v}$ $x\dfrac{\mathrm{d}v}{\mathrm{d}x}=\dfrac{1}{v}$ , so $v\,\mathrm{d}v=\dfrac{\mathrm{d}x}{x}$ $\dfrac{v^{2}}{2}=\ln x+c$ $y=2$ at $x=1$ gives $v=2$ and $c=2$ $y^{2}=2x^{2}\left(\ln x+2\right)$	7
(7 marks)		7
Notes
Dividing numerator and denominator by $x^2$ shows the equation is homogeneous; the substitution reduces it to the separable form $v\,\mathrm{d}v=\tfrac{\mathrm{d}x}{x}$ . Applying the condition after returning to $y$ and $x$ gives $c=2$ . Independent check by differentiating the answer implicitly: $2yy'=4x(\ln x+2)+2x$ , so $y'=\dfrac{2x\ln x+5x}{y}$ , and substituting $y^2=2x^2\ln x+4x^2$ into $\dfrac{x^2+y^2}{xy}$ gives $\dfrac{5x^2+2x^2\ln x}{xy}$ , the same expression.
2	$z=\sin y$ gives $\dfrac{\mathrm{d}z}{\mathrm{d}x}=\cos y\dfrac{\mathrm{d}y}{\mathrm{d}x}$ The equation becomes $\dfrac{\mathrm{d}z}{\mathrm{d}x}+\dfrac{z}{x}=x$ The integrating factor is $x$ , so $\dfrac{\mathrm{d}}{\mathrm{d}x}(xz)=x^{2}$ $xz=\dfrac{x^{3}}{3}+A$ , so $z=\dfrac{x^{2}}{3}+\dfrac{A}{x}$ $\sin y=\dfrac{x^{2}}{3}+\dfrac{A}{x}$	7
(7 marks)		7
Notes
The substitution converts the equation into a first-order linear equation in $z$ with integrating factor $\mathrm{e}^{\int1/x\,\mathrm{d}x}=x$ . Independent check by differentiating the answer: from $x\sin y=\tfrac{x^3}{3}+A$ , differentiating gives $\sin y+x\cos y\,y'=x^2$ , and dividing by $x$ returns the original equation.
3	$u=xy$ gives $\dfrac{\mathrm{d}u}{\mathrm{d}x}=y+x\dfrac{\mathrm{d}y}{\mathrm{d}x}$ $\dfrac{\mathrm{d}^{2}u}{\mathrm{d}x^{2}}=2\dfrac{\mathrm{d}y}{\mathrm{d}x}+x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$ The equation is therefore $\dfrac{\mathrm{d}^{2}u}{\mathrm{d}x^{2}}+u=0$ $u=A\cos x+B\sin x$ $y=\dfrac{A\cos x+B\sin x}{x}$	7
(7 marks)		7
Notes
Differentiating $u=xy$ twice reproduces exactly the combination $xy''+2y'$ , so the equation becomes simple harmonic in $u$ . Independent check by substitution: for $y=\dfrac{\sin x}{x}$ , $xy=\sin x$ , and $\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}(\sin x)+\sin x=0$ , so the original equation is satisfied.
4	$\dfrac{\mathrm{d}u}{\mathrm{d}x}=1+u^{2}$ , which separates as $\dfrac{\mathrm{d}u}{1+u^{2}}=\mathrm{d}x$ $\arctan u=x+c$ , and $u=0$ at $x=0$ gives $c=0$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\tan x$ $y=-\ln\|\cos x\|+d$ , and $y=0$ at $x=0$ gives $d=0$ $y=-\ln(\cos x)$ for $\|x\|<\dfrac{\pi}{2}$	7
(7 marks)		7
Notes
Reducing the order gives a separable equation whose integral is the arctangent; applying the first condition fixes $c=0$ , and integrating $\tan x$ gives $-\ln\|\cos x\|$ . Independent check by differentiation: $y=-\ln(\cos x)$ gives $y'=\tan x$ and $y''=\sec^2x=1+\tan^2x=1+\left(y'\right)^2$ .
5	The equation becomes $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}u^{2}}-4\dfrac{\mathrm{d}y}{\mathrm{d}u}+4y=\mathrm{e}^{2u}$ The auxiliary equation $m^{2}-4m+4=0$ has the repeated root $m=2$ , so the complementary function is $(A+Bu)\mathrm{e}^{2u}$ Because $2$ is a repeated root, try $y=Cu^{2}\mathrm{e}^{2u}$ , giving $2C\mathrm{e}^{2u}=\mathrm{e}^{2u}$ and $C=\dfrac12$ $y=\left(A+Bu+\dfrac{u^{2}}{2}\right)\mathrm{e}^{2u}$ $y=x^{2}\left(A+B\ln x+\dfrac{(\ln x)^{2}}{2}\right)$	8
(8 marks)		8
Notes
Using $x\tfrac{\mathrm{d}y}{\mathrm{d}x}=y_u$ and $x^2\tfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=y_{uu}-y_u$ turns the left side into $y_{uu}-4y_u+4y$ , while $x^2=\mathrm{e}^{2u}$ . The repeated root forces a $u^2$ factor in the particular integral. Independent check by substitution: for $y=\tfrac12x^2(\ln x)^2$ , $y'=x\ln x+x(\ln x)^2$ and $y''=1+3\ln x+(\ln x)^2$ , so $x^2y''-3xy'+4y=x^2\left[1+3\ln x+(\ln x)^2-3\ln x-3(\ln x)^2+2(\ln x)^2\right]=x^2$ .