Change of Variables

**divinelogos** · February 16th 2011, 06:11 PM

I'm having trouble rewriting differential equations in the form F(v). That is, changing a function to make it's variables have the form "y/x" so I can perform the substitution v=y/x. Here's a sample problem:

Solve the differential equation:

yy'= sqrt(x^2+y^2)-x

Is there a general way to go about transforming this into the form f(v)?

Any help would be much appreciated.

Thank!

**alexmahone** · February 16th 2011, 06:18 PM

Originally Posted by divinelogos

I'm having trouble rewriting differential equations in the form F(v). That is, changing a function to make it's variables have the form "y/x" so I can perform the substitution v=y/x. Here's a sample problem:

Solve the differential equation:

yy'= sqrt(x^2+y^2)-x

Is there a general way to go about transforming this into the form f(v)?

Any help would be much appreciated.

Thank!

$\displaystyle yy'=\sqrt{x^2+y^2}-x$

$\displaystyle y'=\frac{\sqrt{x^2+y^2}}y}-\frac{x}{y}$

$\displaystyle y'=\sqrt{\frac{x^2}{y^2}+1}-\frac{x}{y}$

**divinelogos** · February 16th 2011, 06:44 PM

How did you get from the second step to the third step?

**alexmahone** · February 16th 2011, 06:47 PM

Originally Posted by divinelogos

How did you get from the second step to the third step?

$\displaystyle \frac{\sqrt{x^2+y^2}}{y}=\sqrt\frac{x^2+y^2}{y^2}= \sqrt{\frac{x^2}{y^2}+1}$

**divinelogos** · February 16th 2011, 06:49 PM

THAT'S what it was. Some sneaky Algebra right there...Thank you my man!

**divinelogos** · February 16th 2011, 07:18 PM

One more question. That gets it in x/y but I need y/x...

**alexmahone** · February 16th 2011, 07:23 PM

Originally Posted by divinelogos

One more question. That gets it in x/y but I need y/x...

You can still substitute $\displaystyle v=\frac{y}{x}$ so that $\displaystyle \frac{x}{y}=\frac{1}{v}$ .

**divinelogos** · February 16th 2011, 07:54 PM

That works, but it leads to an integral that can only be solved in terms of other integrals. Do you know how they got this?

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**topsquark** · February 16th 2011, 08:58 PM

$\displaystyle y'=\frac{\sqrt{x^2+y^2}}{y}$

$\displaystyle y'= \frac{\sqrt{x^2+y^2}}{ \sqrt{y^2} } = \sqrt{ \frac{x^2 + y^2}{y^2}}$

$\displaystyle = \sqrt{\frac{x^2}{y^2} + \frac{y^2}{y^2}}$

etc.

-Dan

**divinelogos** · February 16th 2011, 09:22 PM

That gets me to:

y'=sqrt((x^2/y^2)) - x/y

I still don't see how they got to that second step :/

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