Solve Seperable diff eq with substitution

[Asphyxiated]

Homework Statement

Solve the differential equation:

x \frac {dy}{dx} = y + e^{\frac {y}{x}}

with the change of variable:

v = \frac {y}{x}

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:

\frac {dy}{dx} = \frac {y}{x} + \frac {e^{\frac{y}{x}}}{x}

then if:

v = \frac {y}{x}

then:

\frac {dv}{dx} = \frac {x*\frac{dy}{dx} - y}{x^{2}}= \frac {\frac{dy}{dx}}{x}-\frac{y}{x^{2}}

so:

\frac {dy}{dx}= \frac{dv}{dx}x+\frac {y}{x}=\frac{dv}{dx}x+v

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

\frac {dv}{dx}x+v=v+\frac{e^{v}}{x}

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

\frac{dv}{dx}x=\frac{e^{v}}{x}

or

\frac{dv}{dx} = \frac {e^{v}}{x^{2}}

this can be separated like so:

\frac {dv}{e^{v}}=\frac {dx}{x^{2}}

\int e^{-v}dv= \int x^{-2}dx

-e^{-v}=-\frac {1}{x} + c

e^{-v}=\frac {1}{x} - c

-v = ln ( \frac {1}{x} -c)

v = -ln ( \frac {1}{x} -c)

then to get the y of the original equation:

\frac {y}{x} = -ln ( \frac {1}{x} -c)

y = -xln ( \frac {1}{x} -c)

 

[Char. Limit]

Looks good to me.

 

[HallsofIvy]

Homework Statement

Solve the differential equation:

x \frac {dy}{dx} = y + e^{\frac {y}{x}}

with the change of variable:

v = \frac {y}{x}

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:

\frac {dy}{dx} = \frac {y}{x} + \frac {e^{\frac{y}{x}}}{x}

then if:

v = \frac {y}{x}

then:

\frac {dv}{dx} = \frac {x*\frac{dy}{dx} - y}{x^{2}}= \frac {\frac{dy}{dx}}{x}-\frac{y}{x^{2}}

You might find it simpler to write that y= xv so that y'= xv'+ v
Then
y'= xv'+ v= v+ \frac{e^v}{x}
xv'= \frac{e^v}{x}
e^{-v}dv= \frac{dx}{x^2}

so:

\frac {dy}{dx}= \frac{dv}{dx}x+\frac {y}{x}=\frac{dv}{dx}x+v

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

\frac {dv}{dx}x+v=v+\frac{e^{v}}{x}

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

\frac{dv}{dx}x=\frac{e^{v}}{x}

or

\frac{dv}{dx} = \frac {e^{v}}{x^{2}}

this can be separated like so:

\frac {dv}{e^{v}}=\frac {dx}{x^{2}}

\int e^{-v}dv= \int x^{-2}dx

-e^{-v}=-\frac {1}{x} + c

e^{-v}=\frac {1}{x} - c

-v = ln ( \frac {1}{x} -c)

v = -ln ( \frac {1}{x} -c)

then to get the y of the original equation:

\frac {y}{x} = -ln ( \frac {1}{x} -c)

y = -xln ( \frac {1}{x} -c)