UPSC 2024 Maths Optional Paper 1 Q8a — Step-by-Step Solution

15 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

Using Laplace transform, solve the initial value problem

y''+2y'+5y=\delta(t-2),\quad y(0)=0,\quad y'(0)=0,

where $\delta(t-2)$ denotes the Dirac delta function.

Standard Laplace transform; complete-the-square inversion; time-shift theorem for the $e^{-2s}$ factor.

Step 1 — Laplace transform.

With zero initial conditions and $\mathcal L\{\delta(t-2)\}=e^{-2s}$ :

(s^2+2s+5)Y(s)=e^{-2s}\;\Rightarrow\;Y(s)=\frac{e^{-2s}}{s^2+2s+5}.

Step 2 — Invert $1/(s^2+2s+5)$ .

Complete the square: $s^2+2s+5=(s+1)^2+4$ .

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

Step 3 — Time-shift theorem.

Since $Y(s)=e^{-2s}\cdot F(s)$ with $f(t)=\tfrac{1}{2}e^{-t}\sin 2t$ :

y(t)=u(t-2)\cdot f(t-2)=u(t-2)\cdot\tfrac{1}{2}e^{-(t-2)}\sin\!\bigl(2(t-2)\bigr).

Explicitly:

\boxed{\;y(t)=u(t-2)\cdot\tfrac{1}{2}e^{-(t-2)}\sin\!\bigl(2(t-2)\bigr).\;}