MHB How to Solve This ODE with Substitution?

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To solve the ODE \(2xyy'=-x^2-y^2\), one effective method is to rewrite it as an exact differential equation, leading to \((x^{2} + y^{2})\ dx + 2\ x\ y\ d y =0\). Another approach involves expressing the equation in the form \(\frac{dy}{dx}=-\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)\) and using the substitution \(v=\frac{y}{x}\). This substitution simplifies the equation and can facilitate finding a solution. Both methods highlight the importance of recognizing the structure of the ODE for effective resolution. Understanding these techniques is crucial for tackling similar equations in the future.
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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:

$$v=\frac{y}{x}$$
 

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