Homogeneous Problem

1. Oct 8, 2006

The Bob

Hi all,

I have been given this question:

Find the initial value problem of the homogeneous equation:

$$(x^2 - y^2) y' = xy \ , \ y(1) = 1$$

Now I know, from my lessons, I have to get it in the form of:

$$\frac{dy}{dx} = f(\frac{y}{x})$$

I have managed to get close but nothing is working out.

If I make the substitution x = vy then I can get $$y' = \frac{v}{1 - v^2}$$ but at this state I have made the substitution twice, not once. I then get to a place where I can seperate the variables:

$$\int \frac{1 - v^2}{v^3} dv = \int \frac{1}{x} dx$$

$$\int \frac{1}{v^3} dv - \int \frac{1}{v} dv = \int \frac{1}{x} dx$$

$$\frac{-1}{2v^2} - ln \vert v \vert = ln \vert x \vert + c$$

Then from here my calculations go all funny. Does anyone know if I am barking up the correct tree?

Cheers,

2. Oct 11, 2006

HallsofIvy

Staff Emeritus
You mean find the solution to this initial value problem!

Actually, if you made that substitution, then the right hand side would be $\frac{v}{v^2-1}$. But you are making the wrong substitution- you don't want to substitute for the independent variable x, you want to substitute for the dependent variable y. The reason for that is that your
$$y'= \frac{v}{1- v^2}[/itex] has only replaced x/y by v on the right side: the left side, y', is still differentiation with respect to x, not v. Let y= vx so that v= y/x. Divide both sides of the equation by $x^2- y^2$ to get $y'= \frac{xy}{x^2- y^2}$ and then divide both numerator and denominator by x2 to get the right hand side to be [tex]\frac{\frac{y}{x}}{1- \left(\frac{y}{x})^2}= \frac{v}{1-v^2}$$
But since y= xv, y'= xv'+ v so your equation is really $xv'+ v= \frac{v}{1- v^2}$ and so $xv'= \frac{v}{1-v^2}-v= \frac{v^3}{1-v^2}$. That is a separable equation and can be solved for v.

3. Oct 12, 2006

The Bob

Cheers HallsofIvy. I didn't think anyone was going to reply. :rofl:

I cannot see how, from $$\frac{-1}{2v^2} - ln|v| = ln|x| + c$$, I can be solved for v. I say that because there is a term in the logarithm and one out of it. Unfortunately, I have not covered this so I am still stuck. I do eventually get to $$y^{2 - y^2} = x^2$$ but I cannot do anything with that either.

Thanks for all your help. I did allow me to see how I should have made the substitution.

Cheers

4. Oct 12, 2006

The Bob

This is wrong. I got to $$\frac{e^{y^2}}{y^{y^2}} = e^{x^2}$$.

5. Oct 13, 2006

HallsofIvy

Staff Emeritus
I don't see how you could get

Cheers,