← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q6b — Step-by-Step Solution

15 marks · Section B

Linear ODE with constant coefficients · ODEs · asked 4× in 14 yrs · Read the full method →

Question

Solve completely $\dfrac{d^2 y}{dx^2}+(\tan x-3\cos x)\dfrac{dy}{dx}+2y\cos^2 x=\cos^4 x$, demonstrating all steps.

Technique

Substitution $t=\sin x$ converts non-constant-coefficient to constant-coefficient $y''-3y'+2y=1-t^2$; standard CF + polynomial PI.

Solution

Strategy. This is a non-constant-coefficient linear ODE. Try a substitution to simplify. Let $t=g(x)$ for some $g$ to make the ODE constant-coefficient.

Step 1 — Recognise potential substitution

Coefficient of $y'$ has $\tan x-3\cos x$; coefficient of $y$ has $\cos^2 x$; RHS has $\cos^4 x$.

Try $t=\sin x$, so $dt/dx=\cos x$.

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot\cos x$.

$\dfrac{d^2 y}{dx^2}=\dfrac{d}{dx}\!\left(\cos x\dfrac{dy}{dt}\right)=-\sin x\dfrac{dy}{dt}+\cos x\cdot\dfrac{d}{dx}\dfrac{dy}{dt}=-\sin x\dfrac{dy}{dt}+\cos^2 x\dfrac{d^2 y}{dt^2}$.

Substitute into the ODE:

$[-\sin x\,y_t+\cos^2 x\,y_{tt}]+(\tan x-3\cos x)\cos x\,y_t+2y\cos^2 x=\cos^4 x$.

Expand: $-\sin x\,y_t+\cos^2 x\,y_{tt}+\sin x\,y_t-3\cos^2 x\,y_t+2\cos^2 x\,y=\cos^4 x$.

The $\sin x\,y_t$ terms cancel: $\cos^2 x\,y_{tt}-3\cos^2 x\,y_t+2\cos^2 x\,y=\cos^4 x$.

Divide by $\cos^2 x$: $y_{tt}-3y_t+2y=\cos^2 x=1-\sin^2 x=1-t^2$.

Step 2 — Solve constant-coefficient ODE

$y''-3y'+2y=1-t^2$ (where $'=d/dt$).

CF: Roots of $D^2-3D+2=(D-1)(D-2)$: $D=1,2$. $y_c=C_1 e^t+C_2 e^{2t}$.

PI for $1-t^2$: Try $y_p=at^2+bt+c$.

$y_p'=2at+b$, $y_p''=2a$.

$2a-3(2at+b)+2(at^2+bt+c)=1-t^2$,

$2at^2+(2b-6a)t+(2a-3b+2c)=1-t^2$.

Match:

• $t^2$: $2a=-1\Rightarrow a=-1/2$.
• $t^1$: $2b-6a=0\Rightarrow 2b+3=0\Rightarrow b=-3/2$.
• $t^0$: $2a-3b+2c=1\Rightarrow -1+9/2+2c=1\Rightarrow 2c=1+1-9/2=-5/2\Rightarrow c=-5/4$.

$y_p=-\dfrac{t^2}{2}-\dfrac{3t}{2}-\dfrac{5}{4}$.

Step 3 — Convert back to $x$

$t=\sin x$.

$y=C_1 e^{\sin x}+C_2 e^{2\sin x}-\dfrac{\sin^2 x}{2}-\dfrac{3\sin x}{2}-\dfrac{5}{4}$.

Answer