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Asphyxiated

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## Homework Statement

Solve the differential equation:

[tex] x \frac {dy}{dx} = y + e^{\frac {y}{x}} [/tex]

with the change of variable:

[tex] v = \frac {y}{x} [/tex]

## Homework Equations

## The Attempt at a Solution

I would just like to know if I have successfully solved the problem. Thanks!

First if we divide by x the equation becomes:

[tex] \frac {dy}{dx} = \frac {y}{x} + \frac {e^{\frac{y}{x}}}{x} [/tex]

then if:

[tex] v = \frac {y}{x} [/tex]

then:

[tex] \frac {dv}{dx} = \frac {x*\frac{dy}{dx} - y}{x^{2}}= \frac {\frac{dy}{dx}}{x}-\frac{y}{x^{2}} [/tex]

so:

[tex] \frac {dy}{dx}= \frac{dv}{dx}x+\frac {y}{x}=\frac{dv}{dx}x+v [/tex]

now I can substitute this back into the original differential equation:

[tex] \frac {dv}{dx}x+v=v+\frac{e^{v}}{x} [/tex]

the v on both sides will cancel out with subtraction so you get:

[tex] \frac{dv}{dx}x=\frac{e^{v}}{x} [/tex]

or

[tex] \frac{dv}{dx} = \frac {e^{v}}{x^{2}} [/tex]

this can be separated like so:

[tex] \frac {dv}{e^{v}}=\frac {dx}{x^{2}} [/tex]

[tex] \int e^{-v}dv= \int x^{-2}dx [/tex]

[tex] -e^{-v}=-\frac {1}{x} + c [/tex]

[tex] e^{-v}=\frac {1}{x} - c [/tex]

[tex] -v = ln ( \frac {1}{x} -c) [/tex]

[tex] v = -ln ( \frac {1}{x} -c) [/tex]

then to get the y of the original equation:

[tex] \frac {y}{x} = -ln ( \frac {1}{x} -c) [/tex]

[tex] y = -xln ( \frac {1}{x} -c) [/tex]