Solve this ODE: $y' = \frac{(y-4x)}{(x-y)}$

Question

My goal, as stated in the title, is to solve this ODE:

$$\frac{dy}{dx} = \frac{y-4x}{x-y}$$

I thought I had solved it by following a particular strategy that I learned, as follows:

Let $u = \frac{y}{x}$ and rearrange so that

$$\frac{dy}{dx} = \frac{u-4}{1-u}$$

By the product rule,

$$\frac{du}{dx} = \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2}$$ and we can again rearrange to get

$$u + x\frac{du}{dx} = \frac{dy}{dx}$$

Substituting this into our expression for $\frac{dy}{dx}$ above yields

$$x\frac{du}{dx} = \frac{u^2 - 4}{1 - u}$$ and this is now a separable ODE. Separate:

$$\frac{1-u}{u^2 - 4}du = \frac{1}{x}dx$$

and solve (the left hand side is a little more trouble than the right, but they are both easily integrable):

$$-\frac{1}{4}ln|u-2|-\frac{3}{4}ln|u+2| = ln|x| + C$$

$$\therefore \space \space e^{-\frac{1}{4}ln|u-2|-\frac{3}{4}ln|u+2|} = e^{ln|x| + C}$$

$$\therefore \space \space |u+2||u-2| = c|x|$$

where I have just made a new constant $c = e^{C+1}$. Now, if I have done everything right, I should just be able to substitute back in for $u$, getting the implicit solution:

$$|\frac{y^2}{x^3}-\frac{4}{x}| = c$$

Except that I have an answer key that tells me I should get this instead:

$$|y+2x|^3|y-2x| = c$$

Am I just bad at algebra? How are these two implicit equations the same? If they aren't, where have I gone wrong?

Answer 1

In the section where you state

$$\therefore \space \space e^{-\frac{1}{4}ln|u-2|-\frac{3}{4}ln|u+2|} = e^{ln|x| + C}$$

$$\therefore \space \space |u+2||u-2| = c|x|$$

Going from the first line, you should instead have

$$\frac{1}{|u-2|^{\frac{1}{4}}|u+2|^{\frac{3}{4}}} = c|x| \tag{1}\label{eq1A}$$

Also, a minor point is that $c = e^{C}$, not $c = e^{C + 1}$.