1. ## [SOLVED] DEq; getting weird answer

Hey all,
Solve: x(dy/dx) = y + 2xe^(-y/x)
let v = y/x
v + x(dv/dx) = v + 2e^(-v) two or three steps later gives:
e^vdv = 2(dx)/x integrate and then take ln gives this
v = ln|2ln|x| + C| this is what i am really wondering about
y/x = ln|2ln|x| + C|
y = xln|2ln|x| +C| this is really weird looking, did i do something wrong? Ive done it like five times.

Thanks,

2. Originally Posted by pberardi
Hey all,
Solve: x(dy/dx) = y + 2xe^(-y/x)
let v = y/x
v + x(dv/dx) = v + 2e^(-v) two or three steps later gives:
e^vdv = 2(dx)/x integrate and then take ln gives this
v = ln|2ln|x| + C| this is what i am really wondering about
y/x = ln|2ln|x| + C|
y = xln|2ln|x| +C| this is really weird looking, did i do something wrong? Ive done it like five times.

Thanks,
Looks good. The inner absolute value signs are necessary but can be avoided by expressing this bit as $\ln x^2$. The other absolute value signs should not be there and really the term $(C + \ln x^2)$ should be positive if we are assuming your answer is real valued. But all your steps are right.

So assuming we can take x,y to be real, I would write the answer

$y = x \ln\left(C + \ln x^2 \right).$

HTH

3. Thanks.