CAIE A-Level Further Math A2 2.6 Differential Equations Questions

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: 9]

Find the solution of the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}+3 y=\sin x

for which y=1 when x=0. Give your answer in the form y=f(x).

$\mathrm{e}^{\int 3 \mathrm{~d} x}=\mathrm{e}^{3 x}$
Finds integrating factor.

$\frac{\mathrm{d}}{\mathrm{d} x}\left(y \mathrm{e}^{3 x}\right)=\mathrm{e}^{3 x} \sin x$
Correct form on LHS and attempt
to integrate RHS.

EITHER
$\begin{aligned} \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=-\mathrm{e}^{3 x} \cos x+3 \int \mathrm{e}^{3 x} \cos x \mathrm{~d} x \text { OR } \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=\frac{1}{3} \mathrm{e}^{3 x} \sin x-\frac{1}{3} \int \mathrm{e}^{3 x} \cos x \mathrm{~d} x \end{aligned}$
Integrates by parts once or uses
$\sin x=\frac{e^{\mathrm{i} x}-\mathrm{e}^{-\mathrm{i} x}}{2 \mathrm{i}}.$

EITHER
$\begin{aligned} \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=-\mathrm{e}^{3 x} \cos x+3\left(\mathrm{e}^{3 x} \sin x-3 \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x\right) \text { OR } \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x=\frac{1}{3} \mathrm{e}^{3 x} \sin x-\frac{1}{3}\left(\frac{1}{3} \mathrm{e}^{3 x} \cos x+\frac{1}{3} \int \mathrm{e}^{3 x} \sin x \mathrm{~d} x\right) \end{aligned}$
Integrates by parts again or substitutes $\frac{\mathrm{e}^{\mathrm{i} x}-\mathrm{e}^{-\mathrm{i} x}}{2 \mathrm{i}}=\sin x$ and
$\frac{\mathrm{e}^{\mathrm{i} x}+\mathrm{e}^{-\mathrm{i} x}}{2}=\cos x.$ $y \mathrm{e}^{3 x}=\frac{1}{10} \mathrm{e}^{3 x}(3 \sin x-\cos x)+C$
Must not see i.

$1=-\frac{1}{10}+C$
Finds C. Substitutes into their
expression (must be integrated).

$y=\frac{3}{10} \sin x-\frac{1}{10} \cos x+\frac{11}{10} e^{-3 x}$
Divides through by coefficient of y.

9

[Maximum number: 7]

The variables x and y are related by the differential equation

6 \frac{\mathrm{~d}^{2} x}{\mathrm{~d} t^{2}}+5 \frac{\mathrm{~d} x}{\mathrm{~d} t}+x=t^{2}+10 t+13 .

(a)

Find the general solution for x in terms of t.

[ 6 ]

$6 m^{2}+5 m+1=0 \Rightarrow m=-\frac{1}{2},-\frac{1}{3}$
Auxiliary equation.

$x=A \mathrm{e}^{-\frac{1}{2} t}+B \mathrm{e}^{-\frac{1}{3} t}$
Complementary function. Allow with ' x= ' missing.

$x=p+q t+r t^{2} \Rightarrow x^{\prime}=q+2 r t \Rightarrow x^{\prime \prime}=2 r$
Particular integral and its derivatives.

$12 r+5 q+10 r t+p+q t+r t^{2}=t^{2}+10 t+13$
Substitutes and equates coefficients.

$r=1, \quad q=0, \quad p=1$ $x=A \mathrm{e}^{-\frac{1}{2} t}+B \mathrm{e}^{-\frac{1}{3} t}+t^{2}+1$
Must have x=.

6

(b)

State an approximate solution for large positive values of t.

[ 1 ]

$x=t^{2}+1$

B1FT

Must have ' x= '. Accept ' x is approximately' but do
not accept ' $x \rightarrow{ }^{\prime}$ . FT on their PI.

1

[Maximum number: 7]

Use the substitution z=x+y to find the solution of the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+3 x+3 y}{3 x+3 y-1}

for which y=0 when x=1. Give your answer in the form $a \ln (x+y)+b(x-y)+c=0$ , where a, b and c are constants to be determined.

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{dz}}{\mathrm{d} x}-1$ $\frac{\mathrm{d} z}{\mathrm{~d} x}-1=\frac{1+3 z}{3 z-1}$
Substitutes and derives first order separable equation.

$\frac{\mathrm{d} z}{\mathrm{~d} x}=\frac{1+3 z}{3 z-1}+1=\frac{6 z}{3 z-1}$ $\int \frac{1}{2}-\frac{1}{6} z^{-1} \mathrm{~d} z=\int 1 \mathrm{~d} x$
Separates variables and integrates both sides.

$\frac{1}{2} z-\frac{1}{6} \ln z=x+C \Rightarrow-\frac{1}{2} x+\frac{1}{2} y-\frac{1}{6} \ln (x+y)=C$ $C=-\frac{1}{2}$
Substitutes initial conditions into their expression.

$\frac{1}{6} \ln (x+y)+\frac{1}{2}(x-y)-\frac{1}{2}=0$
OE.

7

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3