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It's my first week dealing with Differential Equations, and I am totally lost regarding the following question. Any help would be very much appreciated!

u(x)is a solution to initial value problem:

$xy'=y-xe^{\frac{y}{x}}$

y(e)=0

a. $u(e^e)=e^e$

b. $u(e^e)=2^e$

c. $u(e^e)=-e^e$

d. $u(e^e)=e^2$

e. $u(e^e)=e^{-e}$

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  • $\begingroup$ Are you supposed to guess the correct result with just a glance or do you compute the solution first and then check the result? In the latter case one could also just have asked for the value of $u(e^e)$. $\endgroup$ – LutzL Jul 11 at 7:30
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You should recognize that the main intermediate expression of your equation is $v=y/x$. Insert that to get everything to contract nicely to $$ v'=\frac{xy'-y}{x^2}=-\frac{e^v}x, $$ which can now be solved as separable ODE.

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