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(\mathrm dy)/(\mathrm dx)+2y=x^2\log x\implies(\mathrm dy)/(\mathrm dx)+\frac2xy=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(\mathrm dy)/(\mathrm dx)+2xy=x^3\log x`

`(\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=\frac14x^4\log x-\frac1{16}x^4+C`

`y=\frac14x^2\log x-\frac1{16}x^2+\frac C{x^2}`