MAT244-2014F > FE

FE4

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**Victor Ivrii**:

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 $y= -x \ln (e^2+\ln x)$.

**Arash Jalili**:

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**:

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)

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