# finding the general solution for a differential equation

I'm practicing for the DE midterm, going through (supposedly) basic questions to practice.

I am now stuck on this question. I am supposed to find the general equation for the given DE. I don't know what I did wrong. I think the approach is correct, but possibly a calculation error? I can't quite figure it out (I've been trying it for at least an hour)

I divided $$x^2$$ to get rid of the coefficient on $$y'$$, and then applied Bernoulli's method. Next, I used the intergating factor method since it seemed appropriate to use. Some calculation followed and I got y(x) = C*x^(9/2) - which not an answer.

Can anyone point out where I got it wrong? Sorry for the bad handwriting in advance. $$x^2y'+2xy=5y^4$$ • it's done. Hope it helps – Subin Park Nov 4 '18 at 6:17

$$x^2y'+2xy=5y^4$$ $$x^2\frac{y'}{y^4}+2x\frac{1}{y^3}=5$$ $$x^2\frac{-1}{3}\left(\frac{1}{y^3}\right)'+2x\frac{1}{y^3}=5$$ Obviously the change of function $$v=\frac{1}{y^3}$$ is better than $$v=y^3$$. $$v=\frac{1}{y^3}\quad\text{leads to}\quad -x^2v'+6xv=15$$ which is a linear ODE, easy to solve : $$v=cx^6+\frac{15}{7x}$$ $$y^3=\frac{1}{cx^6+\frac{15}{7x}}$$
From what I can see, you used $$y={v^3}$$; use instead $$y^3=\frac 1v$$ and you will arrive to something simple since the equation will become $$x^2 v'-6 x v+15=0$$
$$x^2y'+2xy=5y^4$$ It's really hard to read the picture ...and to point out where you made mistakes $$(x^2y)'=\frac {5y^4x^8} {x^8}$$ $$\int \frac {dx^2y}{x^8y^4}=5\int \frac {dx}{x^8}$$ $$\int \frac {dx^2y}{(x^2y)^4}=- \frac {5}{7x^7}+K$$ $$\frac {1}{3(x^2y)^3}= \frac {5}{7x^7}+K$$ $$x= y^3(\frac {15}{7}+Cx^7)$$ $$\implies y^3= \frac x {(\frac {15}{7}+Cx^7)}$$