Question

$x\frac{dy}{dx}+2y=x^2\log x$. its solution is $y{x}^2=\frac{1}{4}{x}^4\log x-\frac{1}{A}{x}^4+c.$
find $\sqrt{A}$?

Answer 1

$x\frac{dy}{dx}+2y=x^2\log x$
$\frac{dy}{dx}+\frac{2}{x}y=x\log x$
which is a linear differential equation
Here, $P=\frac{2}{x};Q=x\log x$
Integrating factor $I.F.=e^{\int Pdx}$
$I.F=e^{\int \frac{2}{x}dx}$
$\Rightarrow I.F=e^{2\log x}=x^2$
Solution of differential eqn is given by
$y{x}^2=\int x^3\log xdx$
$y{x}^2=\log x\frac{x^4}{4}-\int \frac{1}{x}\frac{x^4}{4}dx$
$\Rightarrow y{x}^2=\frac{x^4}{4}\log x-\frac{1}{4}\int x^{3}dx$
$\Rightarrow y{x}^2=\frac{x^4}{4}\log x-\frac{x^4}{16}+C$
Comparing with given solution , we get
$A=16$
$\Rightarrow \sqrt{A}=4$