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UPSC 2022 Maths Optional Paper 1 Q7b — 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

Solve $y''-3y'+2y=h(t)$ via Laplace transform, where $h(t)=\begin{cases}2,&0<t<4\\ 0,&t>4\end{cases}$, $y(0)=0,\,y'(0)=0$.

Technique

Standard Laplace transform with Heaviside step source; partial fractions; second shifting theorem.

Solution

Setup. $h(t)=2-2u(t-4)$ where $u(\cdot)$ is the Heaviside step.

Laplace: $\mathcal L\{h(t)\}=\dfrac{2}{s}-\dfrac{2}{s}e^{-4s}=\dfrac{2(1-e^{-4s})}{s}$.

Step 1 — Take Laplace transform of ODE

$\mathcal L\{y''\}=s^2 Y-sy(0)-y'(0)=s^2 Y$. $\mathcal L\{y'\}=sY-y(0)=sY$. $\mathcal L\{y\}=Y$.

ODE in $s$-domain: $s^2 Y-3sY+2Y=\dfrac{2(1-e^{-4s})}{s}$.

$Y(s^2-3s+2)=\dfrac{2(1-e^{-4s})}{s}$,

$Y=\dfrac{2(1-e^{-4s})}{s(s^2-3s+2)}=\dfrac{2(1-e^{-4s})}{s(s-1)(s-2)}$.

Step 2 — Partial fractions on $\dfrac{1}{s(s-1)(s-2)}$

$\dfrac{1}{s(s-1)(s-2)}=\dfrac{A}{s}+\dfrac{B}{s-1}+\dfrac{C}{s-2}$.

• $s=0$: $1/[(0-1)(0-2)]=1/2=A$.
• $s=1$: $1/[1\cdot(1-2)]=-1=B$.
• $s=2$: $1/[2\cdot 1]=1/2=C$.

So $\dfrac{1}{s(s-1)(s-2)}=\dfrac{1/2}{s}-\dfrac{1}{s-1}+\dfrac{1/2}{s-2}$.

Step 3 — Inverse transform of “constant” part

$\mathcal L^{-1}\!\left\{\dfrac{2}{s(s-1)(s-2)}\right\}=2\!\left[\dfrac{1/2}{s}-\dfrac{1}{s-1}+\dfrac{1/2}{s-2}\right]\xrightarrow{\mathcal L^{-1}}1-2e^t+e^{2t}=(e^t-1)^2$. Hmm let me redo.

$2\cdot\dfrac{1/2}{s}\xrightarrow{}1$. $2\cdot\dfrac{-1}{s-1}\xrightarrow{}-2e^t$. $2\cdot\dfrac{1/2}{s-2}\xrightarrow{}e^{2t}$.

Sum: $1-2e^t+e^{2t}=(1-e^t)^2$? Check: $(1-e^t)^2=1-2e^t+e^{2t}$ ✓.

So the “constant” part contributes $f(t):=1-2e^t+e^{2t}$.

Step 4 — Inverse transform of $e^{-4s}$ part

$\mathcal L^{-1}\!\left\{e^{-4s}\cdot F(s)\right\}=f(t-4)\cdot u(t-4)$ (second shifting theorem), where $f$ is the inverse of $F$.

$\mathcal L^{-1}\!\left\{\dfrac{2 e^{-4s}}{s(s-1)(s-2)}\right\}=f(t-4)u(t-4)=[1-2e^{t-4}+e^{2(t-4)}]u(t-4)$.

Step 5 — Combine (note the $-$ sign in $Y$)

$Y=\dfrac{2}{s(s-1)(s-2)}-\dfrac{2e^{-4s}}{s(s-1)(s-2)}$.

Answer