Toronto Math Forum
MAT2442014F => MAT244 MathTests => FE => Topic started by: Victor Ivrii 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 $y= x \ln (e^2+\ln x)$.

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?

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)