How can I solve this Bernoulli differential equation? $ 3xy’ - 3xy^4\ln x -y = 0 $

Question

$ 3xy’ - 3xy^4\ln x -y = 0 $ I tried to use $ v= y^{1-n} $ but it wasn’t helpful or at least I couldn’t lead it into a right way.

Answer 1

we write$$-\frac{\frac{3dy(x)}{dx}}{y(x)^4}+\frac{1}{xy(x)^3}=-3\log(x)$$ substituting $$v(x)=\frac{1}{y(x)^3}$$ so we get $$\frac{dv(x)}{dx}+\frac{v(x)}{x}=-3\log(x)$$ with $$\mu=e^{\int\frac{1}{x}dx}=x$$ we get $$x\frac{dv(x)}{dx}+v(x)=-3x\log(x)$$ then we get $$\frac{d}{dx}\left(xv(x)\right)=-3x\log(x)$$ $$v(x)=\frac{3}{4}x-\frac{3}{2}x\log(x)+\frac{C_1}{x}$$ The solution is given by $$y \left( x \right) ={\frac {\sqrt [3]{-4\,x \left( 6\,{x}^{2}\ln \left( x \right) -3\,{x}^{2}-4\,{\it \_C1} \right) ^{2}}}{6\,{x}^{2} \ln \left( x \right) -3\,{x}^{2}-4\,{\it \_C1}}} $$

Answer 2

The identification as Bernoulli DE is correct. Divide by $-y^4$ to get $$ x(y^{-3})'+3x\ln x+(y^{-3})=0 $$ which is now a linear DE in $v=y^{-3}$.

Answer 3

It's Bernouilli's equation $$3xy’ - 3xy^4\ln x -y = 0$$ $$3x\frac {y’}{y^4} -\frac 1{y^3} = 3x\ln x $$ Substitute $z=1/y^3$ $$-xz' -z = 3x\ln x $$ $$xz' +z = -3x\ln x $$ Integrate both sides $$xz = -3\int x\ln xdx $$ Note that $$I=\int x\ln xdx=\frac {x^2\ln |x|}2-\frac 12\int xdx+C=\frac 12x^2( \ln |x|- \frac 12)+C$$ $$z = \frac 32x( -\ln |x|+ \frac 12)+\frac Cx $$ $$\frac 1 {y^3(x)} = \frac 32x( -\ln |x|+ \frac 12)+\frac Cx $$