UPSC 2024 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 the integral equation $y(t)=\cos t+\displaystyle\int_0^t y(x)\cos(t-x)\,dx$ using Laplace transform.

Recognise the integral as a convolution; take the Laplace transform; solve algebraically; invert via complete-the-square.

Step 1 — Laplace transform.

The integral is the convolution $(y\ast\cos)(t)$ . Using $\mathcal L\{f\ast g\}=F(s)G(s)$ and $\mathcal L\{\cos t\}=s/(s^2+1)$ :

Y(s)=\frac{s}{s^2+1}+Y(s)\cdot\frac{s}{s^2+1}.

Step 2 — Solve for $Y(s)$ .

Y(s)\left[1-\frac{s}{s^2+1}\right]=\frac{s}{s^2+1}\;\Rightarrow\;Y(s)\cdot\frac{s^2-s+1}{s^2+1}=\frac{s}{s^2+1}\;\Rightarrow\;Y(s)=\frac{s}{s^2-s+1}.

Step 3 — Inverse Laplace.

Complete the square: $s^2-s+1=(s-\tfrac{1}{2})^2+\tfrac{3}{4}$ . Split the numerator $s=(s-\tfrac{1}{2})+\tfrac{1}{2}$ :

Y(s)=\frac{s-1/2}{(s-1/2)^2+3/4}+\frac{1/2}{(s-1/2)^2+3/4}.

Using standard inversions with $a=1/2,\,\omega=\sqrt{3}/2$ :

\mathcal L^{-1}\!\left\{\frac{s-1/2}{(s-1/2)^2+3/4}\right\}=e^{t/2}\cos\!\tfrac{\sqrt{3}\,t}{2},\qquad \mathcal L^{-1}\!\left\{\frac{1/2}{(s-1/2)^2+3/4}\right\}=\frac{1}{\sqrt{3}}\,e^{t/2}\sin\!\tfrac{\sqrt{3}\,t}{2}.

\boxed{\;y(t)=e^{t/2}\!\left[\cos\!\frac{\sqrt{3}\,t}{2}+\frac{1}{\sqrt{3}}\sin\!\frac{\sqrt{3}\,t}{2}\right].\;}