$\displaystyle xy'-2x^{2}\sqrt{y}=4y$
this equation in not saparable so i need to find the integration coefficient?
is there easier way
it looks like bernuly but i am not sure
because of the square root
If by "integration coefficient" you mean "integrating factor", that technique only works on first-order linear equations, which this obviously is not. Why not try the Bernoulli approach? Standard form would be
$\displaystyle xy'-2x^{2}\sqrt{y}=4y$
$\displaystyle xy'-4y=2x^{2}\sqrt{y}$
$\displaystyle y'-\dfrac{4}{x}\,y=2x\sqrt{y}=2xy^{1/2}.$
So the substitution would be what?
Setting $\displaystyle z=\sqrt{y} \implies dy=2\ z\ dz$ You obtain...
$\displaystyle \displaystyle z^{'} = 2\ \frac{z}{x} +x$ (1)
... that is a linear first order ODE and can be 'attacked' in usual way...
Kind regards
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