Question

Solution of the differential equation $y+\frac{d}{dx}\left(xy\right)=x(\sin x+\ln |x|),$ is:
(where $c$ is integration constant)

• A. $y=\cos x+\frac{2}{x}\sin x+\frac{2}{x^2}\sin x+\frac{x}{3}\ln |x|-\frac{x}{9}+\frac{c}{x^2}$
• B. $y=-\sin x+\frac{2}{x}\cos x+\frac{2}{x^2}\cos x+\frac{x}{3}\ln |x|-\frac{x}{9}+\frac{c}{x^2}$
• C. $y=-\cos x+\frac{2}{x}\sin x+\frac{2}{x^2}\cos x+\frac{x}{3}\ln |x|-\frac{x}{9}+\frac{c}{x^2}$
• D. $y=-\cos x+\frac{3}{x}\sin x+\frac{4}{x^2}\cos x+\frac{x}{3}\ln |x|-\frac{x}{9}+\frac{c}{x^2}$

Answer 1

The correct option is C $y=-\cos x+\frac{2}{x}\sin x+\frac{2}{x^2}\cos x+\frac{x}{3}\ln |x|-\frac{x}{9}+\frac{c}{x^2}$

The given differential equation is
$y+\frac{d}{dx}\left(xy\right)=x(\sin x+\ln |x|)$
i.e., $x\frac{dy}{dx}+2y=x(\sin x+\ln |x|)$
$\Rightarrow \frac{dy}{dx}+\frac{2}{x}y=\sin x+\ln |x|\cdots \left(i\right)$
This is a linear differential equation
On comparing, $\frac{dy}{dx}+Py=Q,$
We get, $P=\frac{2}{x}$ &$Q=\sin x+\ln |x|$
$I.F.=e^{\int \frac{2}{x}dx}=e^{2\ln |x|}=x^2\cdots \left(ii\right)$

$\therefore$ Solution is given by
$y{x}^2=\int x^2(\sin x+\ln |x|)dx+c$
$y{x}^2=-x^2\cos x+2x\sin x+2\cos x+\frac{x^3}{3}\ln |x|-\frac{x^3}{9}+c$
$\Rightarrow y=-\cos x+\frac{2}{x}\sin x+\frac{2}{x^2}\cos x+\frac{x}{3}\ln |x|-\frac{x}{9}+\frac{c}{x^2}.$