# Differential Eq. (Substitution)

1. Feb 20, 2008

1. The problem statement, all variables and given/known data
Solve by making an appropriate substitution. I am given the homogeneous DE:

$$xdx+(y-2x)dy=0$$

Now we have bee using either y=ux or x=vy. . . I tried both, but the latter seemed easier.

$$x\frac{dx}{dy}+y-2x=0$$ letting x=vy and dx/dy=v+y*dy/dv

$$vy(v+y\frac{dy}{dv})+y-2vy=0$$

$$v^2+y^2\frac{dy}{dv}+y-2vy=0$$

Here is where I get stumped. . . this is supposed to be separable now right? Because I can't seem to see it.

A hint would be swell!

Last edited: Feb 20, 2008
2. Feb 21, 2008

### Mathdope

I think you need to check that again.

3. Feb 21, 2008

### sutupidmath

$$x\frac{dx}{dy}+y-2x=0$$, lets divide by x to get

$$\frac{dx}{dy}+\frac{y}{x}=2$$ or

$$\frac{dx}{dy}+(\frac{x}{y})^{-1}=2$$ now lets take the sub

$$v= \frac{x}{y}, x=vy, \frac{dx}{dy}=v+y\frac{dv}{dy}$$,(note: you made a mistake here) now let's go back and substitute we get

$$v+y\frac{dv}{dy}+\frac{1}{v}=2$$ so we get

$$y\frac{dv}{dy}+\frac{v^{2}+1}{v}=2$$

$$y\frac{dv}{dy}=\frac{2v-v^{2}-1}{v}=-\frac{(v-1)^{2}}{v}$$

Now this is separable and i think you will be fine from here on, right?