CAIE A-Level Further Mathematics 2.6.4 Linear Differential Equations

Practise forming complementary functions from auxiliary roots, selecting a suitable particular-integral trial and matching coefficients to obtain a general solution.

Exam points

solve the auxiliary equation and choose the complementary function for repeated or complex roots
select a polynomial, exponential or sine-cosine trial and differentiate it before substitution
equate coefficients to find the particular integral, then add it to the complementary function

[Maximum number: 7]

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} .

Find the general solution for y in terms of t.

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

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

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2.6.4—Linear differential equations question 1