← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Particular integral via operator method · ODEs · asked 7× in 13 yrs · Read the full method →

Question

Solve $\dfrac{d^2 y}{dx^2}+2y=x^2 e^{3x}+e^x\cos 2x$.

Technique

Standard linear ODE; CF from $D^2+2=0$; PI via operator shift for $e^{ax}P(x)$ and $1/(D^2-\alpha^2)$ trick for trig.

Solution

Operator: $f(D)=D^2+2$.

Step 1 — Complementary function

Roots: $D^2+2=0\Rightarrow D=\pm i\sqrt 2$.

$y_c=C_1\cos(\sqrt 2 x)+C_2\sin(\sqrt 2 x)$.

Step 2 — Particular integral for $x^2 e^{3x}$

Use $\dfrac{1}{f(D)}e^{ax}P(x)=e^{ax}\dfrac{1}{f(D+a)}P(x)$ (shift).

$\dfrac{1}{D^2+2}x^2 e^{3x}=e^{3x}\dfrac{1}{(D+3)^2+2}x^2=e^{3x}\dfrac{1}{D^2+6D+11}x^2$.

Expand $1/(D^2+6D+11)$ as a series in $D$ (since $D$ applied to polynomial reduces it): $\dfrac{1}{11(1+(6D+D^2)/11)}=\dfrac{1}{11}\!\left[1-\dfrac{6D+D^2}{11}+\!\left(\dfrac{6D+D^2}{11}\right)^2-\cdots\right]$.

Apply to $x^2$:

• $1\cdot x^2=x^2$.
• $-(6D+D^2)/11 \cdot x^2=-(12x+2)/11$.
• $+(6D+D^2)^2/121\cdot x^2$: $(6D+D^2)^2=36D^2+12D^3+D^4$. On $x^2$: $36\cdot 2+0+0=72$. So $+72/121$.

Higher terms vanish on $x^2$.

Sum: $\dfrac{1}{11}\!\left[x^2-\dfrac{12x+2}{11}+\dfrac{72}{121}\right]=\dfrac{x^2}{11}-\dfrac{12x+2}{121}+\dfrac{72}{1331}$.

$=\dfrac{x^2}{11}-\dfrac{12x}{121}-\dfrac{2}{121}+\dfrac{72}{1331}$.

Common denominator $1331=11^3$: $\dfrac{121 x^2}{1331}-\dfrac{132 x}{1331}-\dfrac{22}{1331}+\dfrac{72}{1331}=\dfrac{121 x^2-132 x+50}{1331}$.

So $y_{p,1}=e^{3x}\cdot\dfrac{121 x^2-132 x+50}{1331}=\dfrac{e^{3x}(121 x^2-132 x+50)}{1331}$.

Step 3 — Particular integral for $e^x\cos 2x$

$\dfrac{1}{D^2+2}e^x\cos 2x=e^x\dfrac{1}{(D+1)^2+2}\cos 2x=e^x\dfrac{1}{D^2+2D+3}\cos 2x$.

On $\cos 2x$: $D^2\to -4$. So $D^2+2D+3\to -4+2D+3=2D-1$.

$\dfrac{1}{2D-1}\cos 2x=\dfrac{1}{2D-1}\cdot\dfrac{2D+1}{2D+1}\cos 2x=\dfrac{(2D+1)\cos 2x}{4D^2-1}$.

$D\cos 2x=-2\sin 2x$, so $(2D+1)\cos 2x=-4\sin 2x+\cos 2x$.

$4D^2-1$ on $\cos 2x$: $4(-4)-1=-17$.

So $\dfrac{1}{2D-1}\cos 2x=\dfrac{-4\sin 2x+\cos 2x}{-17}=\dfrac{4\sin 2x-\cos 2x}{17}$.

Thus $y_{p,2}=\dfrac{e^x(4\sin 2x-\cos 2x)}{17}$.

Step 4 — General solution

Answer