Differential Eq. (Substitution)

[Saladsamurai]

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!

[Mathdope]

dx/dy=v+y*dy/dvI think you need to check that again.

[sutupidmath]

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!

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?