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I'm a little confused about GRE 8767 problem #40.

Let $y=f(x)$ be a solution of the differential equation $xdy+(y-xe^x)dx=0$ such that $y=0$ when $x=1$. What is the value of $f(2)$?

I see that this is an exact differential since $g(x,y)=xy+e^x-xe^x+C$ is such that $g_x=y-xe^x$ and $g_y=x$. How can I use this to find $f(2)$? I tried writing $ g(2,f(2))=2f(2)+e^2-2e^2+C=\cdots $ but I'm not sure what follows. Thanks. The final answer is $e^2/2$ by the way.

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You have arrived at the conclusion that for a solution $y=f(x)$ we have: $ g(x, f(x)) = C $ Initial condition $f(1) = 0$ determines $C=0$, hence $ 0 = g(2, f(2)) = 2 f(2) + \mathrm{e}^2 - 2 \mathrm{e}^2 = 2 f(2) - \mathrm{e}^2 $ Hence $f(2) = \frac{1}{2} \mathrm{e}^2$

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    I do not understand it. Why the downvote?2012-08-28
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Re-write the ODE by multiplying through by $dx$ to give $x \, y'(x) + y(x) - xe^x = 0$. This ODE can be solved very simply. We want to solve $xy' + y = xe^x$ which is the same as $(xy)' = xe^x$. Integrating both sides with respect to $x$ gives $xy = (x-1)e^x + k$ and so:

$y(x) = \frac{x-1}{x}e^x + \frac{k}{x} \, ,$

where $k$ is an arbitrary constant. If $y(1) = 0$ then $k = 0.$ The specific solution is

$y(x) = \frac{x-1}{x}e^x \, , $

and so $y(2) = \frac{1}{2}e^2.$