Question

Differential Eq:
Find the general solution of the following ordinary differential equations (your answer will involve an
unknown constant):
x^2*y' = x*y + x*e^(y/x)

Answer 1

\[x^2*y' = x*y + x*e ^{\frac{ y }{ x }}\]

Answer 2

I'm kind of lost on how to approach this.

Answer 3

What have you tried?

Answer 4

I tried separating it first, but I can't or don't know how I would cancel out e without getting y' stuck in the ln, and if I did that I would to apply e to both sides getting y/x stuck again.

Answer 5

Just verified that its homogenous.

Answer 6

In cases where you see something horrifying like \(e^{y(x) / x} \) it's often a good idea to define a new variable \( u(x) = y(x)/x\) and see how that goes.

Answer 7

have u tried any substituion of variables

Answer 8

I'm guessing I would have to set y=vx?

Answer 9

say u = y/x

Answer 10

y' = v+xv
y'\[y' = \frac{ xy + x^2e^y/x }{ x^2 }\]
\[v +xv' =\frac{ vx^2 +x^2e^v }{ x^2 } = v+e^v\]
\[xv' = e^v\]
\[\frac{ dv }{ dx } = \frac{ e^v }{ x }\]

Answer 11

Then just separate and integrate
Awesome. Thanks, err body

Answer 12

Y'all are the real MVPs.

Answer 13

That's it xD

Answer 14

\[\large y' = \frac{ xy + x^{\color{Red}{2}}e^{y/x} }{ x^2 }\]

double check that exponent, your original equation has just \(x\)

Answer 15

Oh damn. I missed that exponent in the original equation

Answer 16

either ways you will end up wid a separable eq'n, so not to wry too much.. just fix that mistake..

Answer 17

thanks for looking out though.

Answer 18

For fun: Is it possible to solve this with Lambert's W function? =)

Answer 19

who does math for fun! eww nurd

Answer 20

Lambert W function says that x = W(x)*exp(W(x)) right?

Answer 21

Yeah that's right.

Answer 22

then we have 1 = W'(x)*exp(W(x)) + W(x)*W'(x)*exp(W(x)) = (1+W(x))*W'(x)*exp(W(x))

Answer 23

and we have:
v' = exp(v)/x, so we could say that v = W(x), then:
\[\frac{1}{(1+W(x))e^{W(x)}} = \frac{e^{W(x)}}{W(x)e^{W(x)}}\]
\[W(x) = e^{W(x)} + x\]

Answer 24

is that right?

Answer 25

\[W^2(x) = x + xW(x)\]
\[W(x) = \frac{x \pm \sqrt{x^2 + 4x}}{2}\]

Answer 26

Mas W(x) = v(x) = y(x)/x, so:
\[y(x) = \frac{x^2 \pm \sqrt{x^4 + 4x^2}}{2}\]