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UPSC 2014 Maths Optional Paper 1 Q8c — Step-by-Step Solution

20 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 initial value problem

by using Laplace-transform.

Technique

Laplace transform with shift theorem; partial fractions in two irreducible quadratics; complete the square; inverse Laplace via shift theorem.

Solution

Step 1 — Apply Laplace transform

With $Y(s)=\mathcal L\{y(t)\}$ and zero ICs, $\mathcal L\{y''\}=s^{2}Y$.

For RHS: $\mathcal L\{\sin t\}=1/(s^{2}+1)$, so by the shift theorem,

The transformed ODE:

Step 2 — Partial fractions

Expanding and matching coefficients of $s^{3},s^{2},s^{1},s^{0}$:

• $s^{3}$: $A+C=0$.
• $s^{2}$: $4A+B+D=0$.
• $s$: $5A+4B+C=0$.
• $s^{0}$: $5B+D=8$.

Solving: from $C=-A$ in the $s$-equation, $4A+4B=0$, so $A=-B$. Substituting into $s^{2}$-equation: $-4B+B+D=0$, $D=3B$. From $s^{0}$: $5B+3B=8$, $B=1$. Hence $A=-1,\;C=1,\;D=3$.

Step 3 — Inverse Laplace transform

First fraction:

Second fraction: Complete the square in the denominator: $s^{2}+4s+5=(s+2)^{2}+1$. Rewrite the numerator $s+3=(s+2)+1$:

By the shift theorem $\mathcal L^{-1}\{F(s+2)\}=e^{-2t}f(t)$:

Combined: $e^{-2t}(\cos t+\sin t)$.

Step 4 — Sum

Answer