UPSC 2026 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Laplace transform applied to IVP for second-order linear ODE with constant coefficients · ODEs · asked 11× in 14 yrs · Read the full method →

Question

Apply Laplace transform to solve $(D^2 + D)\,y(t) = 2$ , $D \equiv \dfrac{d}{dt}$ , subject to the conditions $y(0) = 3$ , $y'(0) = 1$ .

Technique

The method is the question. Transform the whole equation, using $L\{y''\}=s^2\bar y-sy(0)-y'(0)$ and $L\{y'\}=s\bar y-y(0)$ so that the initial data enter algebraically; solve for $\bar y(s)$ , split by partial fractions, and invert term by term. Do not solve it as a CF + PI problem — that scores nothing here.

Solution

Step 1 — Write the equation and the transform pair.

The equation is

\frac{d^2y}{dt^2} + \frac{dy}{dt} = 2,\qquad y(0)=3,\quad y'(0)=1 .

Write $\bar y(s)=L\{y(t)\}=\displaystyle\int_0^\infty e^{-st}y(t)\,dt$ . The standard derivative rules are

L\{y'(t)\} = s\bar y(s) - y(0),\qquad L\{y''(t)\} = s^2\bar y(s) - s\,y(0) - y'(0),

and for the right-hand side, $L\{2\} = \dfrac{2}{s}$ (valid for $s>0$ ).

Step 2 — Transform, inserting the initial conditions.

Applying $L$ to both sides:

\bigl[s^2\bar y - s\,y(0) - y'(0)\bigr] + \bigl[s\bar y - y(0)\bigr] = \frac{2}{s}.

Putting $y(0)=3$ , $y'(0)=1$ :

\bigl[s^2\bar y - 3s - 1\bigr] + \bigl[s\bar y - 3\bigr] = \frac{2}{s},

(s^2+s)\,\bar y = \frac{2}{s} + 3s + 4 = \frac{3s^2+4s+2}{s}.

Hence, since $s^2+s=s(s+1)$ ,

\bar y(s) = \frac{3s^2+4s+2}{s^2(s+1)} . \tag{$\ast$}

Step 3 — Partial fractions.

The denominator has a double pole at $s=0$ and a simple pole at $s=-1$ , so write

\frac{3s^2+4s+2}{s^2(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} \quad\Longleftrightarrow\quad 3s^2+4s+2 = A\,s(s+1) + B(s+1) + C\,s^2 .

Put $s=0$ : $2 = B$ .
Put $s=-1$ : $3-4+2 = C$ , so $C=1$ .
Compare coefficients of $s^2$ : $3 = A + C$ , so $A=2$ .

Therefore

\bar y(s) = \frac{2}{s} + \frac{2}{s^2} + \frac{1}{s+1}.

(Quick check of the split: $\dfrac{2s(s+1)+2(s+1)+s^2}{s^2(s+1)} = \dfrac{3s^2+4s+2}{s^2(s+1)}$ ✓.)

Step 4 — Invert.

Using $L^{-1}\{1/s\}=1$ , $L^{-1}\{1/s^2\}=t$ , $L^{-1}\{1/(s+a)\}=e^{-at}$ :

y(t) = 2 + 2t + e^{-t}.

Step 5 — Check against the data and the equation.

$y(0)=2+0+1=3$ ✓. $y'(t)=2-e^{-t}$ , so $y'(0)=2-1=1$ ✓. $y''(t)=e^{-t}$ , so

y''+y' = e^{-t} + 2 - e^{-t} = 2 \ \checkmark

identically in $t$ .

(Structural remark, one line in the exam: the double pole at $s=0$ is what produces the secular term $2t$ — physically the particular integral of a constant forcing for an operator $D(D+1)$ with a zero root; the simple pole at $s=-1$ gives the transient $e^{-t}$ .)

Answer

\boxed{\;y(t) = 2 + 2t + e^{-t}\;}