UPSC 2016 Maths Optional Paper 1 Q6d — Step-by-Step Solution

10 marks · Section B

Laplace transform applied to IVP for second-order linear ODE with constant coefficients · ODEs · asked 11× in 14 yrs · Read the full method →

Question

Using Laplace transformation, solve the following: $y''-2y'-8y=0,\ y(0)=3,\ y'(0)=6$ .

Laplace transform turns the IVP into algebra: $(s^2-2s-8)Y=3s$ ; partial fractions over $(s-4)(s+2)$ ; invert term by term.

Let $Y(s)=\mathcal L\{y\}$ . Using $\mathcal L\{y'\}=sY-y(0)$ and $\mathcal L\{y''\}=s^2Y-sy(0)-y'(0)$ with $y(0)=3,\ y'(0)=6$ :

\big[s^2Y-3s-6\big]-2\big[sY-3\big]-8Y=0.

(s^2-2s-8)Y-3s-6+6=0\ \Rightarrow\ (s^2-2s-8)Y=3s.

Factor $s^2-2s-8=(s-4)(s+2)$ :

Y(s)=\frac{3s}{(s-4)(s+2)}.

\frac{3s}{(s-4)(s+2)}=\frac{A}{s-4}+\frac{B}{s+2}.

$3s=A(s+2)+B(s-4)$ . At $s=4$ : $12=6A\Rightarrow A=2$ . At $s=-2$ : $-6=-6B\Rightarrow B=1$ . So

Y(s)=\frac{2}{s-4}+\frac{1}{s+2}.

Using $\mathcal L^{-1}\{1/(s-a)\}=e^{at}$ ,

\boxed{\;y(x)=2e^{4x}+e^{-2x}.\;}