Question

Find the general solution of the differential equation $x\frac{dy}{dx}+2y=x^{2}logx.$

Answer 1

We have,

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

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

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

Comparing this equation

$\frac{dy}{dx}+Py=Q$

Then, $P=\frac{2}{x},Q=x\log x$

$I.F.=e^{\int Pdx}$

$=e^{\int \frac{2}{x}dx}$

$=e^{2\log x}$

$=e^{{\mathrm{logx}}^2}\therefore e^{\log x}=x$

$=x^2$

Then, Complete solution is

$yI.F.=\int I.F.Qdx+C$

$y{x}^2=\int x^2.x\log xdx+C$

$y{x}^2=\int x^3\log xdx+C$

Now,

$\int x^3\log xdx=\log x\int x^{3}dx-\int \left(\frac{d\log x}{dx}{x}^{3}dx\right)dx+C$

$=\frac{x^4\log x}{4}-\int x^{2}dx+C$

$=\frac{x^4\log x}{4}-\frac{x^3}{3}+C$

$y{x}^2$ = $\frac{x^4\log x}{4}-\frac{x^3}{3}+C$

$y=$ $\frac{x^{}\log x}{4}-\frac{x}{3}+C$

Hence, this is the answer.