$\begingroup$I am not sure what am I doing wrong here, I have $$x^2y'+2xy=y^3$$ I should apply the Bernoulli equation, so I mark $$z=\frac{1}{y^2}$$ $$z' = -\frac{2y'}{y^3}$$ $$-\frac{z'}{2} = \frac{y'}{y^3}$$
Now I just substitute $z$ and I get $$-\frac{x^2z'}{2}+2xz=1$$
How I should proceed from here? I am not recognize any pattern
$\endgroup$13 Answers
$\begingroup$$$x^2y\prime +2xy=y^3$$ Rewrite the DE as: $$(x^2y)'=y^3$$ $$\dfrac {d(x^2y)}{(x^2y)^3}=\dfrac {dx}{x^6}$$ Integrate.
$\endgroup$$\begingroup$From that step you can write your equation in the form $$\frac{dz}{dx}+Pz=Q$$ where $P$ and $Q$ depend on $x$...
You get $$\frac{dz}{dx}-\frac{4}{x}z=-2x^{-2}$$
Your $P=-\frac{4}{x}$, $Q=-2x^{-2}$
Now, $$ze^{\int-\frac{4}{x}dx}=\int{e^{\int-\frac{4}{x}dx}(-2x^{-2})dx}$$
So that $$zx^{-4}=-2\int{x^{-6}dx}$$.
$\endgroup$$\begingroup$Your last equation in $z$ is
$$x^2z'-4xz=-2$$ without RHS,
$$\frac{z'}{z}=\frac{4x}{x^2}=\frac 4x$$ then $$\ln(\frac{z}{\lambda})=\ln(x^4)$$ and
$$z=\lambda.x^4$$ let us look for a particular solution as$$z_p=\lambda(x)x^4$$ with
$$\lambda'x^6=-2$$ thus $$\lambda=\frac{-2}{5x^5}$$ therefore $$z=\lambda.x^4-\frac{2}{5x}$$ finally
$$\Large{y=\sqrt{\frac{1}{\lambda.x^4-\frac{2}{5x}}}}$$ where $\lambda\in \Bbb R$.
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