Question in Differential Equation. (Don't know how to proceed) By using the substitution $y=vx$, find the particular solution of the differential equation $$2xy\frac{dy}{dx}=y^2-x^2$$, given that y=4 when x=2. Express in terms of x.
My attempt,
$$y=xv$$
$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$
$$2x^2(x\frac{dv}{dx}+v)-x^2+x^2v^2$$
$$2x^2(x\frac{dv}{dx}+v)v=x^2(v^2-1)$$
$$\frac{dv}{dx}=\frac{-v^2-1}{2xv}$$
$$\frac{2\frac{dv}{dx}v}{-v^2-1}=\frac{1}{x}$$
$$\int \frac{2\frac{dv}{dx}v}{-v^2-1}=\int \frac{1}{x}dx$$
$$-ln (v^2+1)=\ln (x)+c_1$$
How do I simplify this? 

here is a third way, using integrating factor,  to solve this differential equation. write the differential equation as $$M\,dx + N\, dy = (x^2 - y^2)\, dx + 2xy \, dy = 0 $$ this differential equation is not exact because $M_y = (x^2 - y^2)_y = -2y \neq N_x = (2xy)_x=2y$ but almost.  we will look for a function, called an inegrating fator, $\mu$ so that $$\mu (x^2 - y^2)\, dx + 2\mu xy \, dy = 0 $$ is exact. the condition is $$ -2\mu y + (x^2 - y^2)\mu_y=\mu(x^2 - y^2)_y + (x^2 - y^2)\mu_y = (\mu2xy)_x=2y\mu + 2xy \mu_x$$ that is $$(x^2 - y^2)\mu_y - 2xy\mu_x = 4\mu y\tag 1$$ now, choose $\mu$ a function of $x$ alone, this turns the partial differential equation into an ordinary differential equation 
$$\frac{d\mu}{\mu} = -\frac{2dx}{x} \to \mu = \frac 1{x^2}.$$  therefore 
$$ 0=\frac{(x^2 - y^2)}{x^2}\, dx + \frac{2y}{x^2} \, dy = dx + d\left(\frac{y^2}{x}\right)$$ giving the integral $$x + \frac{y^2}{x} = 2 + \frac{16}2 = 10$$ and the solution is $$y = \sqrt{10x - x^2}. $$
