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
y&#039;= \frac{v}{1- v^2}[/itex]<br /> has only replaced x/y by v on the <b>right</b> side: the left side, y&#039;, is still differentiation with respect to <b>x</b>, not v.<br /> <br /> Let y= vx so that v= y/x. Divide both sides of the equation by x^2- y^2 to get y&amp;#039;= \frac{xy}{x^2- y^2} and then divide both numerator and denominator by x² to get the right hand side to be <br /> \frac{\frac{y}{x}}{1- \left(\frac{y}{x})^2}= \frac{v}{1-v^2}<br /> But since y= xv, y&#039;= xv&#039;+ v so your equation is really xv&amp;#039;+ v= \frac{v}{1- v^2} and so xv&amp;#039;= \frac{v}{1-v^2}-v= \frac{v^3}{1-v^2}. That is a separable equation and can be solved for v.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I then get to a place where I can separate the variables:<br /> <br /> \int \frac{1 - v^2}{v^3} dv = \int \frac{1}{x} dx<br /> <br /> \int \frac{1}{v^3} dv - \int \frac{1}{v} dv = \int \frac{1}{x} dx<br /> <br /> \frac{-1}{2v^2} - ln \vert v \vert = ln \vert x \vert + c<br /> <br /> Then from here my calculations go all funny. Does anyone know if I am barking up the correct tree?<br /> <br /> Cheers,<br /> <br /> The Bob (2004 ©) </div> </div> </blockquote>

 

[The Bob]

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

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
y^{y^2}[/itex]<br /> <br /> 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
y^{y^2}[/itex]<br /> <br /> 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.

<br /> <br /> 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.<br /> <br /> Cheers,<br /> <br /> The Bob (2004 ©)