← 2020 Paper 1

UPSC 2020 Maths Optional Paper 1 Q7b — Step-by-Step Solution

10 marks · Section B

Properties of Laplace transform (linearity, shift, derivative, convolution) · ODEs · asked 3× in 13 yrs · Read the full method →

Question

Using Laplace transform, solve the initial value problem ty′′+2ty′+2y=2ty''+2ty'+2y=2; y(0)=1y(0)=1 and y′(0)y'(0) is arbitrary. Does this problem have a unique solution?

Technique

Laplace transform with the L{tf}=−ddsL{f}\mathcal{L}\{t f\}=-\tfrac{d}{ds}\mathcal{L}\{f\} rule turns the ODE into a first-order linear ODE in Y(s)Y(s); solve with an integrating factor, partial-fraction, invert.

Solution

Let Y(s)=L{y(t)}Y(s)=\mathcal{L}\{y(t)\}, y(0)=1y(0)=1, y′(0)=ay'(0)=a (a number we leave free). Recall

L{y′}=sY−y(0),L{y′′}=s2Y−sy(0)−y′(0),\mathcal{L}\{y'\}=sY-y(0),\quad \mathcal{L}\{y''\}=s^2Y-sy(0)-y'(0), L{t f(t)}=−ddsL{f(t)}.\mathcal{L}\{t\,f(t)\}=-\frac{d}{ds}\mathcal{L}\{f(t)\}.

Step 1 — Transform each term

L{ty′′}=−dds(s2Y−s−a)=−(2sY+s2Y′−1),\mathcal{L}\{ty''\}=-\frac{d}{ds}\big(s^2Y-s-a\big)=-\big(2sY+s^2Y'-1\big), L{ty′}=−dds(sY−1)=−(Y+sY′),\mathcal{L}\{ty'\}=-\frac{d}{ds}\big(sY-1\big)=-\big(Y+sY'\big), L{2y}=2Y,L{2}=2s.\mathcal{L}\{2y\}=2Y,\qquad \mathcal{L}\{2\}=\frac{2}{s}.

Step 2 — Assemble the subsidiary equation

−(2sY+s2Y′−1)+2[−(Y+sY′)]+2Y=2s.-\big(2sY+s^2Y'-1\big)+2\big[-(Y+sY')\big]+2Y=\frac{2}{s}. −s2Y′−2sY+1−2sY′−2Y+2Y=2s  ⇒  −(s2+2s)Y′−2sY=2s−1.-s^2Y'-2sY+1-2sY'-2Y+2Y=\frac{2}{s} \;\Rightarrow\;-(s^2+2s)Y'-2sY=\frac{2}{s}-1 .

Multiply by −1-1:

(s2+2s)Y′+2sY=1−2s.(∗)(s^2+2s)Y'+2sY=1-\frac{2}{s}. \tag{$\ast$}

Note the y′(0)=ay'(0)=a term cancelled out — the initial slope leaves no trace in (∗)(\ast). This already signals non-uniqueness.

Step 3 — Solve the linear first-order ODE in YY

Write (∗)(\ast) as

Y′+2ss(s+2)Y=1−2/ss(s+2)  ⇒  Y′+2s+2Y=s−2s2(s+2).Y'+\frac{2s}{s(s+2)}Y=\frac{1-2/s}{s(s+2)} \;\Rightarrow\;Y'+\frac{2}{s+2}Y=\frac{s-2}{s^2(s+2)} .

Integrating factor: μ=e∫2s+2ds=(s+2)2\mu=e^{\int \frac{2}{s+2}ds}=(s+2)^2. Then

dds[(s+2)2Y]=(s+2)2⋅s−2s2(s+2)=(s+2)(s−2)s2=s2−4s2=1−4s2.\frac{d}{ds}\big[(s+2)^2Y\big]=(s+2)^2\cdot\frac{s-2}{s^2(s+2)}=\frac{(s+2)(s-2)}{s^2}=\frac{s^2-4}{s^2}=1-\frac{4}{s^2}.

Integrate:

(s+2)2Y=s+4s+C1.(s+2)^2Y=s+\frac{4}{s}+C_1 .

So

Y=s+4s+C1(s+2)2=s2+C1s+4s(s+2)2.Y=\frac{s+\tfrac{4}{s}+C_1}{(s+2)^2}=\frac{s^2+C_1 s+4}{s(s+2)^2}.

Step 4 — Partial fractions

s2+C1s+4s(s+2)2=1s+C1−4(s+2)2.\frac{s^2+C_1 s+4}{s(s+2)^2}=\frac{1}{s}+\frac{C_1-4}{(s+2)^2}.

(Check: 1s\tfrac{1}{s} gives (s+2)2s(s+2)2=s2+4s+4s(s+2)2\tfrac{(s+2)^2}{s(s+2)^2}=\tfrac{s^2+4s+4}{s(s+2)^2}; adding C1−4(s+2)2=(C1−4)ss(s+2)2\tfrac{C_1-4}{(s+2)^2}=\tfrac{(C_1-4)s}{s(s+2)^2} yields numerator s2+4s+4+(C1−4)s=s2+C1s+4s^2+4s+4+(C_1-4)s=s^2+C_1 s+4 ✓.)

Let C=C1−4C=C_1-4 (arbitrary constant). Then

Y(s)=1s+C(s+2)2.Y(s)=\frac{1}{s}+\frac{C}{(s+2)^2}.

Step 5 — Inverse transform

L−1{1s}=1,L−1{1(s+2)2}=t e−2t.\mathcal{L}^{-1}\Big\{\tfrac1s\Big\}=1,\qquad \mathcal{L}^{-1}\Big\{\tfrac{1}{(s+2)^2}\Big\}=t\,e^{-2t}.  y(t)=1+C t e−2t,C arbitrary \boxed{\ y(t)=1+C\,t\,e^{-2t},\quad C\ \text{arbitrary}\ }

Step 6 — Uniqueness

Answer

 No, the solution is not unique. \boxed{\ \text{No, the solution is not unique.}\ }
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