UPSC 2020 Maths Optional Paper 1 Q6a — Step-by-Step Solution

20 marks · Section B

Method of variation of parameters · ODEs · asked 11× in 13 yrs · Read the full method →

Question

Using the method of variation of parameters, solve the differential equation $y''+(1-\cot x)y'-y\cot x=\sin^2 x$ , if $y=e^{-x}$ is one solution of CF.

Technique

Factor the operator as $(D-\cot x)(D+1)$ to get $y_2=\sin x-\cos x$ ; then Wronskian-based variation of parameters with $R=\sin^2 x$ .

Solution

The equation is $y''+(1-\cot x)y'-(\cot x)\,y=\sin^2 x$ , standard form with

P(x)=1-\cot x,\quad Q(x)=-\cot x,\quad R(x)=\sin^2 x .

Step 1 — Find the second CF solution

The given solution is $y_1=e^{-x}$ . The operator factors neatly:

(D-\cot x)(D+1)y=(D-\cot x)(y'+y)=y''+y'-\cot x\,(y'+y)=y''+(1-\cot x)y'-\cot x\,y .

So the homogeneous equation is $(D-\cot x)(D+1)y=0$ . Put $v=(D+1)y$ ; then $(D-\cot x)v=0\Rightarrow \dfrac{dv}{v}=\cot x\,dx\Rightarrow v=\sin x$ . Solving $(D+1)y=\sin x$ for a non- $e^{-x}$ solution:

y'+y=\sin x,\quad \text{IF }e^{x}:\ (e^x y)'=e^x\sin x,\ e^x y=\tfrac{e^x}{2}(\sin x-\cos x),

so a second independent solution is

y_2=\tfrac12(\sin x-\cos x)\ \Rightarrow\ \text{take } y_2=\sin x-\cos x .

Check: with $y_2=\sin x-\cos x$ , $y_2'=\cos x+\sin x$ , $y_2''=-\sin x+\cos x$ :

y_2''+(1-\cot x)y_2'-\cot x\,y_2 =(\cos x-\sin x)+(\cos x+\sin x)-\cot x(\cos x+\sin x)-\cot x(\sin x-\cos x)

=2\cos x-\cot x(2\sin x)=2\cos x-2\cos x=0.\ \checkmark

Complementary function:

y_c=C_1 e^{-x}+C_2(\sin x-\cos x).

Step 2 — Wronskian

W=\begin{vmatrix}y_1 & y_2\\ y_1' & y_2'\end{vmatrix} =e^{-x}(\cos x+\sin x)-(-e^{-x})(\sin x-\cos x)

=e^{-x}\big[(\cos x+\sin x)+(\sin x-\cos x)\big]=e^{-x}\,(2\sin x)=2e^{-x}\sin x .

Step 3 — Variation of parameters

Seek $y_p=u_1y_1+u_2y_2$ with

u_1'=-\frac{y_2 R}{W},\qquad u_2'=\frac{y_1 R}{W}.

$u_2$ :

u_2'=\frac{e^{-x}\sin^2 x}{2e^{-x}\sin x}=\frac{\sin x}{2} \ \Rightarrow\ u_2=-\tfrac12\cos x .

$u_1$ :

u_1'=-\frac{(\sin x-\cos x)\sin^2 x}{2e^{-x}\sin x}=-\frac{e^{x}\sin x(\sin x-\cos x)}{2} =-\frac{e^x}{2}\big(\sin^2 x-\sin x\cos x\big).

Using $\sin^2 x=\tfrac{1-\cos2x}{2}$ and $\sin x\cos x=\tfrac12\sin2x$ :

u_1'=-\frac{e^x}{4}\big(1-\cos2x-\sin2x\big).

Integrate (use $\int e^x\cos2x\,dx=\tfrac{e^x}{5}(\cos2x+2\sin2x)$ , $\int e^x\sin2x\,dx=\tfrac{e^x}{5}(\sin2x-2\cos2x)$ ):

u_1=-\frac14\Big[e^x-\tfrac{e^x}{5}(\cos2x+2\sin2x)-\tfrac{e^x}{5}(\sin2x-2\cos2x)\Big] =-\frac{e^x}{4}\Big[1+\tfrac{1}{5}\cos2x-\tfrac{3}{5}\sin2x\Big]

=\frac{e^x}{20}\big(3\sin2x-\cos2x-5\big).

Step 4 — Particular integral

y_p=u_1y_1+u_2y_2=\frac{1}{20}\big(3\sin2x-\cos2x-5\big)-\tfrac12\cos x(\sin x-\cos x).

Now $-\tfrac12\cos x\sin x=-\tfrac14\sin2x$ and $\tfrac12\cos^2 x=\tfrac14(1+\cos2x)$ , so the second piece $=-\tfrac14\sin2x+\tfrac14+\tfrac14\cos2x$ . Adding:

y_p=\tfrac{3}{20}\sin2x-\tfrac{1}{20}\cos2x-\tfrac14-\tfrac14\sin2x+\tfrac14+\tfrac14\cos2x =-\tfrac{1}{10}\sin2x+\tfrac15\cos2x .

(The constant terms $-\tfrac14+\tfrac14$ cancel.)

y_p=-\frac{1}{10}\sin 2x+\frac{1}{5}\cos 2x .

Step 5 — General solution

Answer

\boxed{\,y=C_1 e^{-x}+C_2(\sin x-\cos x)-\frac{1}{10}\sin 2x+\frac{1}{5}\cos 2x\,}