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:

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

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