Question

For the given differential equation find the general solution.

$x\frac{dy}{dx}+2y=x^{2}logx$

Answer 1

Given, $x\frac{dy}{dx}+2y=x^{2}logx$
On dividing by x on both sides, we get $\frac{dy}{dx}+2\frac{y}{x}=xlogx$
On comparing with the from $\frac{dy}{dx}+Py=Q,$ we get
$P=\frac{2}{x}andQ=xlogx$
$\therefore IF=e^{\int Pdx}\Rightarrow e^{2\int \frac{1}{x}dx}=IF=e^{2log|x|}=x^2$
The general solution of the given differential equation is given by
$y.IF=\int Q\times IFdx+C\Rightarrow x^{2}y=\int x^2.xlogxdx+C\Rightarrow x^{2}y=\int x^{3}logxdx+C\Rightarrow x^{2}y=logx.\int x^3-\int [\frac{d}{dx}logx\int x^{3}dx]dx+C\Rightarrow x^{2}y=logx.\frac{x^4}{4}-\int [\frac{d}{dx}logx\int x^{3}dx]dx+C\Rightarrow x^{2}y=logx.\frac{x^4}{4}-\int [\frac{1}{x}\times \frac{x^4}{4}]dx+C\Rightarrow x^{2}y=logx.\frac{x^4}{4}-\int \frac{x^3}{4}dx+C\Rightarrow x^{2}y=\frac{x^4}{4}logx-\frac{x^4}{16}+C\Rightarrow y=\frac{x^2}{16}(4logx-1)+C{x}^{-2}$