Question

X dy/dx +2y =x^2logx by using Bernoulli's Equation

Answer 1

This ODE isn't of Bernoulli type, but it is linear, so we should be able to find an integrating factor to solve it.

$x\dfrac{\mathrm dy}{\mathrm dx}+2y=x^2\log x\implies\dfrac{\mathrm dy}{\mathrm dx}+\dfrac2xy=x\log x$

The integrating factor will be

$\mu(x)=\exp\left(\displaystyle\int\frac2x\,\mathrm dx\right)=x^2$

Multiplying both sides of the ODE by the IF gives

$x^2\dfrac{\mathrm dy}{\mathrm dx}+2xy=x^3\log x$
$\dfrac{\mathrm d}{\mathrm dx}[x^2y]=x^3\log x$
$x^2y=\displaystyle\int x^3\log x\,\mathrm dx$

Integrate the right hand side by parts to get

$x^2y=\dfrac14x^4\log x-\dfrac1{16}x^4+C$
$y=\dfrac14x^2\log x-\dfrac1{16}x^2+\dfrac C{x^2}$