Find the general solution to the ODE

Question

I would like to find the general solution to the ODE: $$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{(1+x^2)^3(4x+1)}{4-x}y^5$$

and the tips given is "let $z:=\frac{1}{y^4}$. What is the differential equation that z satisfies?"

I tried by sub $$z=\frac{1}{y^4}$$ $$dz=-\frac{4}{y^5}dy$$

but the equation doesn't seem familiar to me. May I know how should I proceed? Thanks.

Answer 1

HINT

Multiply both sides by $1 + x^{2}$ in order to obtain \begin{align*} (1+x^{2})y' + 2xy = \frac{(1+x^{2})^{4}(4x+1)}{4-x}y^{5} \Longleftrightarrow ((1+x^{2})y)' = \frac{4x+1}{(1+x^{2})(4-x)}((1+x^{2})y)^{5} \end{align*}

Then make the substitution $u = (1+x^{2})y$ in order to obtain the following ODE \begin{align*} u' = \frac{(4x+1)u^{5}}{(1+x^{2})(4-x)} \end{align*} which is separable.

Can you take it from here?

Answer 2

$$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{(1+x^2)^3(4x+1)}{4-x}y^5$$ $$(1+x^2)\frac{dy}{dx}+{2x}y=\frac{(1+x^2)^4(4x+1)}{4-x}y^5$$ $$((1+x^2)y)'=\frac{(1+x^2)^4(4x+1)}{4-x}y^5$$ Try to separate now. $$\dfrac {((1+x^2)y)'}{(y(1+x^2))^5}=\frac{(4x+1)}{(4-x)(1+x^2)}$$ Integrate both sides. $$\int \dfrac {d(y(1+x^2))}{(y(1+x^2))^5}=\int \frac{(4x+1)}{(4-x)(1+x^2)} \, dx$$