UPSC Maths 2013 Paper 1 Q6d — Solution

15 marks · Section B

Question

By using Laplace transform method, solve the differential equation

(D^{2}+n^{2})x=a\sin(nt+\alpha),\qquad D^{2}\equiv\frac{d^{2}}{dt^{2}}

subject to the initial conditions $x=0$ and $\dfrac{dx}{dt}=0$ at $t=0$ , in which $a,n,\alpha$ are constants.

Technique

Standard Laplace approach for linear ODEs with constant coefficients and zero IC; the RHS Laplace produces a denominator $(s^{2}+n^{2})$ , which when multiplied by the system transfer denominator yields $(s^{2}+n^{2})^{2}$ — a resonance signature.

Solution

Strategy. Apply Laplace transform; the zero initial conditions simplify the $L\{x''\}$ term; invert via standard pairs.

Step 1 — Apply Laplace transform

With $X(s)=\mathcal L\{x(t)\}$ and $x(0)=x'(0)=0$ :

\mathcal L\{x''\}=s^{2}X(s).

For the RHS, expand $\sin(nt+\alpha)=\sin nt\cos\alpha+\cos nt\sin\alpha$ , so

\mathcal L\{\sin(nt+\alpha)\}=\frac{n\cos\alpha+s\sin\alpha}{s^{2}+n^{2}}.

The transformed ODE:

(s^{2}+n^{2})X(s)=a\cdot\frac{n\cos\alpha+s\sin\alpha}{s^{2}+n^{2}}\;\Longrightarrow\;X(s)=\frac{a(n\cos\alpha+s\sin\alpha)}{(s^{2}+n^{2})^{2}}.

Step 2 — Inverse Laplace via standard pairs

Use these inverse-Laplace results:

\mathcal L^{-1}\!\left\{\frac{1}{(s^{2}+n^{2})^{2}}\right\}=\frac{\sin nt-nt\cos nt}{2n^{3}},\qquad\mathcal L^{-1}\!\left\{\frac{s}{(s^{2}+n^{2})^{2}}\right\}=\frac{t\sin nt}{2n}.

(The first is derived via convolution of $\sin nt/n$ with itself; the second is a standard table entry, equivalent to $\mathcal L\{t\sin nt\}=2ns/(s^{2}+n^{2})^{2}$ .)

Substitute:

x(t)=a\cdot n\cos\alpha\cdot\frac{\sin nt-nt\cos nt}{2n^{3}}+a\sin\alpha\cdot\frac{t\sin nt}{2n}.

Simplify:

x(t)=\frac{a\cos\alpha}{2n^{2}}(\sin nt-nt\cos nt)+\frac{at\sin\alpha\sin nt}{2n}.

Step 3 — Combine using compound-angle formula

Factor out $\dfrac{a}{2n^{2}}$ :

x(t)=\frac{a}{2n^{2}}\!\left[\cos\alpha\sin nt-nt\cos\alpha\cos nt+nt\sin\alpha\sin nt\right].

The middle and last terms combine: $-nt(\cos\alpha\cos nt-\sin\alpha\sin nt)=-nt\cos(nt+\alpha)$ . Therefore

Answer

\boxed{\;x(t)=\dfrac{a}{2n^{2}}\bigl[\cos\alpha\,\sin nt-nt\cos(nt+\alpha)\bigr].\;}