Solving a DE using substitution method. Please Help!

Question

I'm attempting to solve the following DE:

$$y^2\frac{dy}{dx} +\frac{y^3}{x}=\frac{2}{x^2}$$

with the substitution $u(x) = y^3$

I can somewhat picture what needs to be done, but I seem to come up short. For instance, I know the following:

$$u(x) = y^3$$

$$\frac{du}{dx}= 3y^2\frac{dy}{dx}$$

Therefore:

$$\frac{dy}{dx}= \frac{1}{3y^2}\frac{du}{dx}$$

Furthermore:

$$\frac{dy}{dx}=\frac{1}{3u^{\frac{2}{3}}}\frac{du}{dx}$$

Substituting this into the DE is where I'm getting stuck. I can't seem to get it to "work" after that point.

I would certainly appreciate any help!

Answer 1

Multiply your fierst equation by 3 to egt $$\frac{du}{dx}+\frac{3}{x}u=\frac{6}{x^2},$$ Which is a linear DE, using the integrating factor you can solve it as follows $$\beta=\exp{\int \frac{3}{x}dx}=x^3\\ u=x^{-3}\{\int x^3*\frac{6}{x^2}dx+c\}$$

Answer 2

Hint: By what you wrote, we have $$\frac{1}{3}\frac{du}{dx}+\frac{u}{x}=\frac{2}{x^2},$$ or equivalently $$\frac{du}{dx}+\frac{3}{x}u=\frac{6}{x^2}.\tag{1}$$ You will not have any trouble with the equation $\frac{du}{dx}+\frac{3}{x}u=0$. You will also need a particular solution of (1).