### Author Topic: FE4  (Read 2605 times)

#### Victor Ivrii ##### FE4
« on: December 08, 2014, 04:14:27 PM »
Find the general solution of the ODE
\begin{equation*}
x y' = y - x e^{\frac{y}{x}}
\end{equation*}
and solve the initial value problem $\ y(1) = -2\$.

Solution
Since it is homogeneous equation we plug $y=ux$ and then
\begin{equation*}
u'x^2+ux=ux -xe^{u}\implies u'=-e^{u}\implies x^{-1}dx=-e^{-u}du\implies
\ln x = e^{-u}+\ln C\implies u =-\ln \ln (Cx)\implies y=-x\ln \ln (Cx).
\end{equation*}
As $x=1$, $y=-2$, $u=-2$ we get $\ln \ln C=2$, and $y= -x \ln (e^2+\ln x)$.
« Last Edit: December 14, 2014, 03:59:36 PM by Victor Ivrii »

#### Arash Jalili

• Newbie
• • Posts: 3
• Karma: 0 ##### Re: FE4
« Reply #1 on: December 10, 2014, 12:26:57 PM »
I will take a chance since no one has responded. this could be very very wrong.

v = y/x
=> y=vx => y' = v + v'x

xy' = y - xe^(y/x) => y' = y/x - e^(y/x)
=> v + v'x = v - e^v
=> v'x = -e^v
=> (-e^-v)dv = dx/x
=> e^(-v) = ln x + c
=> ln(e^-v) = ln(ln x +c)
=> v = -ln(ln(x +c)
=> y = -xln(ln(x+c)

y(1)=-2 => c = e^-2
=> y = -xln(ln(x + e^-2)

Other than this I could think of using taylor series for the e^(y/x) to make it linear?

#### Arash Jalili

• Newbie
• • Posts: 3
• Karma: 0 ##### Re: FE4
« Reply #2 on: December 10, 2014, 04:40:02 PM »
just realized I had a typo (missing a parenthesis):

v = y/x
=> y=vx => y' = v + v'x

xy' = y - xe^(y/x) => y' = y/x - e^(y/x)
=> v + v'x = v - e^v
=> v'x = -e^v
=> (-e^-v)dv = dx/x
=> e^(-v) = ln (x) + c
=> ln(e^-v) = ln(ln (x) +c)
=> v = -ln(ln(x) +c)
=> y = -xln(ln(x)+c)

y(1)=-2 => c = e^-2
=> y = -xln(ln(x) + e^-2)