CAIE A-Level Further Math 2.6 Differential equations Question Bank

Practise solving first- and second-order differential equations, using substitutions, initial conditions and exact-form answers.

Exam points

choose an integrating factor or auxiliary equation from the form of the differential equation

Covered syllabus points

2.6.1—Linear differential equations 2.6.2—Linear differential equations 2.6.3—Linear differential equations 2.6.4—Linear differential equations 2.6.5—Differential substitutions 2.6.6—Initial conditions

[Maximum number: 11]

(a)

Show that an appropriate integrating factor for

\left(x^{2}+1\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+y \sqrt{x^{2}+1}=x^{2}-x \sqrt{x^{2}+1}

is $x+\sqrt{x^{2}+1}$ .

[ 4 ]

$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{y}{\sqrt{ }\left(x^{2}+1\right)}=\frac{x}{x^{2}+1}\left(x-\sqrt{ }\left(x^{2}+1\right)\right)$

B1

$\mathrm{e}^{\int \frac{1}{\sqrt{\left(x^{2}+1\right)}} \mathrm{d} x}=\mathrm{e}^{\sinh ^{-1} x}$

M1 A1

$=x+\sqrt{ }\left(x^{2}+1\right)$

A1

7a 4

(b)

Hence find the solution of the differential equation

\left(x^{2}+1\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+y \sqrt{x^{2}+1}=x^{2}-x \sqrt{x^{2}+1}

for which $y=\ln 2$ when x=0. Give your answer in the form y=f(x).

[ 7 ]

$\frac{\mathrm{d}}{\mathrm{d} x}\left(y\left(x+\sqrt{ }\left(x^{2}+1\right)\right)\right)=-\frac{x}{x^{2}+1}$

M1 A1

$y\left(x+\sqrt{ }\left(x^{2}+1\right)\right)=-\int \frac{x}{x^{2}+1} \mathrm{~d} x=-\frac{1}{2} \ln \left(x^{2}+1\right)+C$

M1 A1

$\ln 2=C$

M1

$y=\frac{\ln 2-\frac{1}{2} \ln \left(x^{2}+1\right)}{x+\sqrt{ }\left(x^{2}+1\right)}=\left(x-\sqrt{ }\left(x^{2}+1\right)\right) \ln \left(\frac{1}{2} \sqrt{ }\left(x^{2}+1\right)\right)$

M1 A1

7b 7

[Maximum number: 11]

It is given that $x=t^{3} y$ and

t^{3} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} t^{2}}+\left(4 t^{3}+6 t^{2}\right) \frac{\mathrm{d} y}{\mathrm{~d} t}+\left(13 t^{3}+12 t^{2}+6 t\right) y=61 \mathrm{e}^{\frac{1}{2} t} .

(a)

Show that

\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4 \frac{\mathrm{~d} x}{\mathrm{~d} t}+13 x=6 \mathrm{e}^{\frac{1}{2} t}

[ 4 ]

$\frac{\mathrm{d} x}{\mathrm{~d} t}=t^{3} \frac{\mathrm{~d} y}{\mathrm{~d} t}+3 t^{2} y$

B1

$\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=t^{3} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} t^{2}}+6 t^{2} \frac{\mathrm{~d} y}{\mathrm{~d} t}+6 t y$

B1

$\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4 \frac{\mathrm{~d} x}{\mathrm{~d} t}+13 x=t^{3} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} t^{2}}+6 t^{2} \frac{\mathrm{~d} y}{\mathrm{~d} t}+6 t y+4 t^{3} \frac{\mathrm{~d} y}{\mathrm{~d} t}+12 t^{2} y+13 t^{3} y=61 \mathrm{e}^{\frac{1}{2} t}$

M1 A1

7a 4

(b)

Find the general solution for y in terms of t.

[ 7 ]

$m^{2}+4 m+13=0 \Rightarrow m=-2 \pm 3 i$

M1

$x=\mathrm{e}^{-2 t}(A \cos 3 t+B \sin 3 t)$

A1

$x=k \mathrm{e}^{\frac{1}{2} t} \Rightarrow \dot{x}=\frac{1}{2} k \mathrm{e}^{\frac{1}{2} t} \Rightarrow \ddot{x}=\frac{1}{4} k \mathrm{e}^{\frac{1}{2} t}$

B1

$\frac{1}{4} k+2 k+13 k=61 \Rightarrow k=4$

M1 A1

$t^{3} y=\mathrm{e}^{-2 t}(A \cos 3 t+B \sin 3 t)+4 \mathrm{e}^{\frac{1}{2} t} \Rightarrow y=t^{-3} \mathrm{e}^{-2 t}(A \cos 3 t+B \sin 3 t)+4 t^{-3} \mathrm{e}^{\frac{1}{2} t}$

M1 A1

7b 7

All question bank results loaded

Covered syllabus points

2.6 Differential equations question 1

Question (a)

Question (b)

2.6 Differential equations question 2

Question (a)

Question (b)