Solving Homogeneous Problem: Initial Value & Substitution

[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 separate 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,

The Bob (2004 ©)

 

[HallsofIvy]

Hi all,

I have been given this question:

Find the initial value problem of the homogeneous equation:

You mean find the solution to this initial value problem!

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

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.

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
[tex]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
$$\frac{\frac{y}{x}}{1- \left(\frac{y}{x})^2}= \frac{v}{1-v^2}$$
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.

I then get to a place where I can separate 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,

The Bob (2004 ©)

 

[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

The Bob (2004 ©)

 

[The Bob]

$$y^{2 - y^2} = x^2$$

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

The Bob (2004 ©)

 

[HallsofIvy]

I don't see how you could get
[tex]y^{y^2}[/itex]

I get $-\frac{1}{2}\frax{x^2}{y^2}- ln\left(\frac{y}{x}\right)= ln -\frac{1}{2}$. Are you required to solve for y? In general, in problems like this, that is impossible.

 

[The Bob]

I don't see how you could get
[tex]y^{y^2}[/itex]

I get $-\frac{1}{2}\frax{x^2}{y^2}- ln\left(\frac{y}{x}\right)= ln -\frac{1}{2}$. Are you required to solve for y? In general, in problems like this, that is impossible.

I do not know what I got off hand, unfortunately I am in a rush. I do agree that $$y^{y^2}$$ is not correct.

Cheers,

The Bob (2004 ©)