How to solve this Homogeneous differential equation?

Question

$$ \frac{4xy}{(x^2-y^2)}\frac{dy}{dx} = 1$$ when $y=0 , x=1 $ show that $$ \sqrt{x}.(x^2-5y^2) =1 $$ using this substitute $ y=vx $.

Answer 1

$$\frac{4vx^2}{(x^2-v^2x^2)}(v'x+v) = 1$$

is

$$\frac{4v}{(1-v^2)}(v'x+v) = 1$$

or

$$v'x =\frac{1-v^2}{4v}-v=\frac{1-5v^2}{4v}$$ which is separable. After logarithmic integration,

$$1-5v^2=Cx^{-5/2}$$

which is

$$1-5\frac{y^2}{x^2}=\frac C{x^2\sqrt x}.$$

Answer 2

You have to wirte $$y'=u'x+u$$ so you will get $$\frac{4u}{1-u^2}(u'x+u)=1$$ Can you finish? Simplifying we get $$u'x=\frac{1-5u^2}{4u}$$ Now write $$\frac{2}{5}\frac{10u}{1-5u^2}du=\frac{dx}{x}$$ and substitute $$t=1-5u^2$$ then you will get $$\frac{2}{5}\int\frac{10u}{1-5u^2}du=-\frac{2}{5}\ln|5u^2-1|+C$$