solving the Bernoulli differential equation $y'x^2\ln y-y=xy'$

Question

I tried solving the Bernoulli differential equation $y'x^2\ln y-y=xy'$, but I didn't succeed cause I can see no way to bring it to a canonical form....are there other ways to solve it ?

Answer 1

$$y'x^2\ln y-y=xy'$$ Consider $x'=\frac {dx}{dy}=\frac 1 {y'}$ $$ x^2\ln y-yx'=x$$ Then solve for x the Bernouilli's equation $$\implies x'y+x=x^2\ln y$$ Or you can integrate this way.. $$(xy)'=(xy)^2 \frac {\ln y}{y^2}$$ It's separable... $$ \int \frac {d(xy)}{(xy)^2}= \int \frac {\ln y}{y^2} dy$$ For the integral integrate by part $$I=\int \frac {\ln y}{y^2} dy=-\frac {\ln|y|}y+\int \frac {dy}{y^2}$$ $$............$$

Answer 2

Write $$\frac{1}{\frac{dx(y)}{dy}}=\frac{y}{\log(y)x(y)-1)x(y)}$$ so $$-\frac{\frac{dx(y)}{dy}}{x(y)^2}-\frac{1}{yx(y)}=-\frac{\log(y)}{y}$$ substituting $$v(y)=\frac{1}{x(y)}$$ then $$\frac{v(y)}{dy}-\frac{v(y)}{y}=-\frac{\log(y)}{y}$$ with $$\mu(y)=\frac{1}{y}$$ we get $$\frac{\frac{dv(y)}{dy}}{y}+\frac{d}{dy}\left(\frac{1}{y}v(y)\right)=-\frac{\log(y)}{y^2}$$ so $$\int \frac{d}{dy}\left(\frac{v(y)}{y}\right)dy=\int-\frac{\log(y)}{y^2}dy$$ so $$\frac{v(y)}{y}=\frac{1}{y}+\frac{\log(y)}{y}+C_1$$ and we get $$x(y)=\frac{1}{v(y)}=\frac{1}{\log(y)+yC_1+1}$$