Nonlinear ODE with Hint: Solving for x and y

[Kranz]

Hello everyone and thanks for looking at my thread,

I had some trouble solving this ODE which was in a textbook by Henry J. Ricardo:

$$x(e^y - y') = 2$$.

This problem is from a section dealing with linear equations, but there is a hint beside the problem which reads, "Hint: Think of y as the independent variable, x as the dependent variable, and rewrite the equation in terms of dx/dy."

I tried to solve it by doing the following:

$$e^y - y' = 2 / x$$
$$y' = e^y - 2 / x$$
$$y' = (xe^y - 2) / x$$
$$dx/dy = x / (xe^y - 2)$$
$$xe^y + e^y (dx/dy) = x$$
$$dx/dy + x = xe^{-y}$$
$$dx/dy = x(e^{-y} - 1)$$
$$(1 / x)dx = (e^{-y} - 1)dy$$
$$ln|x| + C = -e^{-y} - y$$

However, I know that this is not the correct solution; I'm guessing that I went wrong somewhere when I rewrote the equation in terms of $$dx/dy$$, but I don't know how else I would approach the problem. Where did I go wrong with my work, and how would I proceed to solve the problem correctly?

Thank you very much for your help!

 

[torquil]

I think you made an error when going from line 4 to line 5?. Line 4 starting with dx/dy is ok. But the next line doesn't seem right? I haven't check in detail, only quickly in my head, so who knows...

Torquil

 

[IttyBittyBit]

I didn't look at your solution, but if you want to solve it just substitute e^y = z and it becomes the Bernoulli equation.