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UPSC 2021 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Laplace transform applied to IVP for second-order linear ODE with constant coefficients · ODEs · asked 10× in 13 yrs · Read the full method →

Question

Solve $\dfrac{d^2 y}{dx^2}+4y=e^{-2x}\sin 2x$, $y(0)=y'(0)=0$, using Laplace transform.

Technique

Laplace transform with shift $\mathcal L\{e^{-2x}\sin 2x\}=2/((s+2)^2+4)$; partial fractions; inverse via standard pairs.

Solution

Step 1 — Take Laplace transform

$\mathcal L\{y''\}=s^2 Y-sy(0)-y'(0)=s^2 Y$. $\mathcal L\{y\}=Y$. $\mathcal L\{e^{-2x}\sin 2x\}=\dfrac{2}{(s+2)^2+4}$ (shift theorem: $\mathcal L\{\sin 2x\}=2/(s^2+4)$, shift $s\to s+2$).

Equation: $(s^2+4)Y=\dfrac{2}{(s+2)^2+4}=\dfrac{2}{s^2+4s+8}$.

$Y=\dfrac{2}{(s^2+4)(s^2+4s+8)}$.

Step 2 — Partial fractions

$\dfrac{2}{(s^2+4)(s^2+4s+8)}=\dfrac{As+B}{s^2+4}+\dfrac{Cs+D}{s^2+4s+8}$.

$2=(As+B)(s^2+4s+8)+(Cs+D)(s^2+4)$.

Expand: $(As+B)(s^2+4s+8)=As^3+4As^2+8As+Bs^2+4Bs+8B$ $=As^3+(4A+B)s^2+(8A+4B)s+8B$.

$(Cs+D)(s^2+4)=Cs^3+Ds^2+4Cs+4D$.

Sum: $s^3$: $A+C=0$. $s^2$: $4A+B+D=0$. $s^1$: $8A+4B+4C=0$. $s^0$: $8B+4D=2$.

From $A+C=0$: $C=-A$. Sub into $s^1$: $8A+4B-4A=0\Rightarrow 4A+4B=0\Rightarrow B=-A$.

Sub into $s^0$: $8(-A)+4D=2\Rightarrow -8A+4D=2\Rightarrow D=(2+8A)/4=1/2+2A$.

Sub into $s^2$: $4A+(-A)+(1/2+2A)=0\Rightarrow 5A+1/2=0\Rightarrow A=-1/10$.

Then $B=1/10$, $C=1/10$, $D=1/2-1/5=3/10$.

So $Y=\dfrac{-s/10+1/10}{s^2+4}+\dfrac{s/10+3/10}{s^2+4s+8}$ $=\dfrac{1}{10}\cdot\dfrac{1-s}{s^2+4}+\dfrac{1}{10}\cdot\dfrac{s+3}{s^2+4s+8}$.

Step 3 — Invert

First term: $\dfrac{1-s}{s^2+4}=\dfrac{1}{s^2+4}-\dfrac{s}{s^2+4}\xrightarrow{}\dfrac{\sin 2x}{2}-\cos 2x$.

So $\dfrac{1}{10}\cdot(\dfrac{\sin 2x}{2}-\cos 2x)$.

Second term: Complete square: $s^2+4s+8=(s+2)^2+4$.

$\dfrac{s+3}{(s+2)^2+4}=\dfrac{(s+2)+1}{(s+2)^2+4}=\dfrac{s+2}{(s+2)^2+4}+\dfrac{1}{(s+2)^2+4}$.

Invert: $e^{-2x}\cos 2x+\dfrac{1}{2}e^{-2x}\sin 2x$.

So $\dfrac{1}{10}[e^{-2x}\cos 2x+\dfrac{1}{2}e^{-2x}\sin 2x]$.

Step 4 — Combine

Simplify:

Answer