MHB How to Solve This ODE with Substitution?

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I haven't done ODEs in a while nor have a book handing.

How do I tackle an equation of the form
\[
2xyy'=-x^2-y^2
\]
I tried polar but that didn't seem to work.
 
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Re: separable or not separable ODE

wmccunes said:
I haven't done ODEs in a while nor have a book handing.

How do I tackle an equation of the form
\[
2xyy'=-x^2-y^2
\]
I tried polar but that didn't seem to work.

With simple steps Yoy arrive to write...

$\displaystyle (x^{2} + y^{2})\ dx + 2\ x\ y\ d y =0\ (1)$ ... and the expression (1) is an 'exact differential'... Kind regards $\chi$ $\sigma$
 
Re: separable or not separable ODE

Another approach would be to write the ODE as:

$$\frac{dy}{dx}=-\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x} \right)$$

and use the substitution:

$$v=\frac{y}{x}$$
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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