UPSC 2016 Maths Optional Paper 1 Q6b — Step-by-Step Solution

15 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 $(D^2+2D+1)y=e^{-x}\log(x)$ , $\left[D\equiv\dfrac{d}{dx}\right]$ .

Variation of parameters with $y_1=e^{-x},y_2=xe^{-x}$ , $W=e^{-2x}$ ; the exponentials cancel leaving $\int x\log x\,dx$ and $\int\log x\,dx$ .

$D^2+2D+1=(D+1)^2=0$ gives the repeated root $D=-1$ . Hence

y_c=(c_1+c_2x)e^{-x},\qquad y_1=e^{-x},\ \ y_2=xe^{-x}.

y_1'=-e^{-x},\qquad y_2'=(1-x)e^{-x}.

W=\begin{vmatrix}e^{-x}&xe^{-x}\\ -e^{-x}&(1-x)e^{-x}\end{vmatrix}=e^{-2x}\big[(1-x)+x\big]=e^{-2x}.

With $R(x)=e^{-x}\log x$ (RHS in standard form, leading coefficient $1$ ),

y_p=-y_1\!\int\frac{y_2R}{W}\,dx+y_2\!\int\frac{y_1R}{W}\,dx.

Compute the integrands:

\frac{y_2R}{W}=\frac{xe^{-x}\cdot e^{-x}\log x}{e^{-2x}}=x\log x,\qquad \frac{y_1R}{W}=\frac{e^{-x}\cdot e^{-x}\log x}{e^{-2x}}=\log x.

\int x\log x\,dx=\frac{x^2}{2}\log x-\int\frac{x}{2}\,dx=\frac{x^2}{2}\log x-\frac{x^2}{4},

\int\log x\,dx=x\log x-x.

y_p=-e^{-x}\!\left(\frac{x^2}{2}\log x-\frac{x^2}{4}\right)+xe^{-x}\big(x\log x-x\big).

=e^{-x}\!\left[-\frac{x^2}{2}\log x+\frac{x^2}{4}+x^2\log x-x^2\right]=e^{-x}\!\left[\frac{x^2}{2}\log x-\frac{3x^2}{4}\right].

\boxed{\;y=(c_1+c_2x)e^{-x}+e^{-x}\!\left(\frac{x^2}{2}\log x-\frac{3x^2}{4}\right).\;}