Question

Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

Answer 1

`y'-\frac13y=\frac13xe^x\ln x\,y^(-2)`

Divide both sides by
`\frac13y^(-2)(x)`:

`3y^2y'-y^3=xe^x\ln x`

Substitute
`v(x)=y(x)^3`, so that
`v'(x)=3y(x)^2y'(x)`.

`v'-v=xe^x\ln x`

Multiply both sides by
`e^(-x)`:

`e^(-x)v'-e^(-x)v=x\ln x`

The left side can be condensed into the derivative of a product.

`(e^(-x)v)'=x\ln x`

Integrate both sides to get

`e^(-x)v=\frac12x^2\ln x-\frac14x^2+C`

Solve for
`v(x)`:

`v=\frac12x^2e^x\ln x-\frac14x^2e^x+Ce^x`

Solve for
`y(x)`:

`y^3=\frac12x^2e^x\ln x-\frac14x^2e^x+Ce^x`

`\implies\boxed{y(x)=\sqrt[3]{\frac14x^2e^x(2\ln x-1)+Ce^x}}`