Question

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

Answer 1

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$

Divide both sides by $\dfrac13y^{-2}(x)$:

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

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

$e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C$

Solve for $v(x)$:

$v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

Solve for $y(x)$:

$y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

$\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}$