1. ## Change of Variables

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!

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

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

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

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

4. Originally Posted by divinelogos
How did you get from the second step to the third step?
$\displaystyle \frac{\sqrt{x^2+y^2}}{y}=\sqrt\frac{x^2+y^2}{y^2}= \sqrt{\frac{x^2}{y^2}+1}$

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

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

7. 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 $\displaystyle \frac{x}{y}=\frac{1}{v}$.

8. 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?

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

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

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

etc.

-Dan

10. That gets me to:

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

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