CAIE A-Level Further Math A2 2.6.5 Differential Substitutions Questions

Practise solving differential equations using substitutions that simplify the dependent expression with Further Mathematics Paper 2 questions and mark schemes.

Exam points

Apply de Moivre's theorem to powers and roots of complex numbers.

[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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CAIE A-Level Further Math A2 2.6.5 Differential Substitutions Questions question 1