veksetz

2022-01-20

Some double angle identity to solve $(2{x}^{2}+{y}^{2})\frac{dy}{dx}=2xy?$

Mary Herrera

Beginner2022-01-20Added 37 answers

If we write the equation as,

$\frac{dy}{dx}=\frac{2xy}{2{x}^{2}+{y}^{2}}$

and then divide through$x}^{2$ we will get:

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

This suggests that we simplify the previous equation in terms of

$f\left(v\right)=\frac{2v}{2+{v}^{2}}$

So putting

$\frac{y}{x}=v$

$y=vx$

$y=vx+v$

We get

$\frac{dv}{dx}x+v=\frac{2v}{2+{v}^{2}}$

Then

$\frac{dv}{dx}x=-\frac{{v}^{3}}{2+{v}^{2}}$

$\frac{dx}{x}=-\frac{2+{v}^{2}}{{v}^{3}}dv$

$\frac{dx}{x}=(-\frac{2}{{v}^{3}}-\frac{1}{v})dv$

Upon integration we have:

$\mathrm{log}x+C=\frac{1}{{v}^{2}}-\mathrm{log}v$

Lets

and then divide through

This suggests that we simplify the previous equation in terms of

So putting

We get

Then

Upon integration we have:

Lets

Elaine Verrett

Beginner2022-01-21Added 41 answers

Rewrite equation into form :

$\frac{dy}{dx}=\frac{2xy}{2{x}^{2}+{y}^{2}}$

Substitute:

$z=\frac{y}{x}\Rightarrow y=xz+z$

Therefore:

$xz+z=frac2z2+{z}^{2}Rightarrowxz=\frac{-{z}^{3}}{2+{z}^{2}}\Rightarrow \int \frac{2+{z}^{2}}{{z}^{3}}dz=-\int \frac{dx}{x}$

Substitute:

Therefore:

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I Completed the square on the bottom but what do you do now?

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inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$