UPSC 2025 Maths Optional Paper 1 Q6a — Step-by-Step Solution

15 marks · Section B

Properties of Laplace transform (linearity, shift, derivative, convolution) · ODEs · asked 4× in 14 yrs · Read the full method →

Question

If $F(s)$ and $G(s)$ are Laplace transforms of $f(t)$ and $g(t)$ respectively, then prove that $\mathcal{L}\left(\displaystyle\int_0^t f(x)\,g(t - x)\,dx\right) = F(s)\,G(s)$ . Using this result, solve the equation $y(t) = t + \displaystyle\int_0^t y(x)\sin(t - x)\,dx$ .

Technique

Prove the Laplace convolution theorem, then apply it to the integral equation: transform to the $s$ -domain, solve algebraically for $Y(s)$ , and invert.

Solution

Part 1 — Proof of the convolution theorem

By definition,

\mathcal{L}\!\left(\int_0^t f(x)\,g(t-x)\,dx\right) = \int_0^\infty e^{-st}\left(\int_0^t f(x)\,g(t-x)\,dx\right)dt.

The region of integration is $\{(x,t): 0\le x\le t<\infty\}$ . Interchange the order (Fubini, valid since both transforms exist absolutely):

= \int_0^\infty f(x)\left(\int_x^\infty e^{-st}\,g(t-x)\,dt\right)dx.

In the inner integral put $\tau = t - x$ ( $d\tau = dt$ , $\tau: 0\to\infty$ ):

\int_x^\infty e^{-st} g(t-x)\,dt = \int_0^\infty e^{-s(\tau+x)} g(\tau)\,d\tau = e^{-sx}\int_0^\infty e^{-s\tau} g(\tau)\,d\tau = e^{-sx}G(s).

Therefore

\mathcal{L}\!\left(\int_0^t f(x)g(t-x)\,dx\right) = \int_0^\infty f(x)\,e^{-sx}\,dx \cdot G(s) = F(s)\,G(s). \qquad\blacksquare

Part 2 — Solving the integral equation

y(t) = t + \int_0^t y(x)\sin(t-x)\,dx = t + (y * \sin)(t).

Take Laplace transforms, with $Y(s)=\mathcal{L}\{y\}$ , $\mathcal{L}\{t\}=\dfrac{1}{s^2}$ , $\mathcal{L}\{\sin t\}=\dfrac{1}{s^2+1}$ , and using the convolution theorem on the last term:

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

Solve for $Y$ :

Y\left(1 - \frac{1}{s^2+1}\right) = \frac{1}{s^2} \;\Longrightarrow\; Y\cdot\frac{s^2}{s^2+1} = \frac{1}{s^2} \;\Longrightarrow\; Y(s) = \frac{s^2+1}{s^4} = \frac{1}{s^2} + \frac{1}{s^4}.

Invert termwise, using $\mathcal{L}^{-1}\{1/s^2\}=t$ and $\mathcal{L}^{-1}\{1/s^4\}=\dfrac{t^3}{3!}=\dfrac{t^3}{6}$ :

y(t) = t + \frac{t^3}{6}.

Answer

\boxed{\;\mathcal{L}\!\left(\int_0^t f(x)g(t-x)\,dx\right) = F(s)G(s),\qquad y(t) = t + \frac{t^3}{6}.\;}