Hard differential equation

[kate]

(x² - 4)y' - 4y = -(x + 2)y²

[Guest]

Also you have the Bernoulli eq. and you can start with a substitution
because you can rewrite your eq. as
[tex]\frac{y'}{y^2}-\frac{4}{y(x^2-4)}=-\frac{1}{x-2}[/tex]
and then
[tex]z(x)=\frac{1}{y(x)}[/tex]
[tex]z'=-\frac{1}{y^2}[/tex]
and you get
[tex]z'+\frac{4z}{(x^2-4)}=-\frac{1}{x-2}[/tex]
with help of integration factor
[tex](z*e^{ln\frac{2-x}{x+2}})'=-\frac{1}{x-2}*\frac{2-x}{x+2}[/tex]
[tex]z*\frac{2-x}{x+2}=c-ln(x+2)[/tex]
[tex]z=(c-ln(x+2))\frac{x+2}{2-x}[/tex]
and thus y
[tex]z=\frac{2-x}{(c-ln(x+2))(x+2)}[/tex]

[Guest]

hello everybody
I have a hard PDE to slove. is anybody to help me?
df/dt + udf/dx= sin(kx-wt)dg/du vlasov equation
can i uplode pic?