CAIE A-Level Further Mathematics 2.6.5 Differential Substitutions

Practise differentiating a stated substitution, reducing the original differential equation to a standard form and converting the solved result back to the required variables.

Exam points

differentiate the new variable carefully and replace every derivative in the original equation
separate variables or solve the resulting constant-coefficient linear differential equation
apply any initial condition in the transformed relation before returning to the requested form

[Maximum number: 4]

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} .

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}

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

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

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2.6.5—Differential substitutions question 1