<s>[INST] 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? 
 [/INST] exponentiating $$\ln (v^2 + 1) = -\ln x + \ln C, $$ you get 
$$ v^2 + 1 = \frac C x \to \frac{y^2 }{x^2} + 1=\frac C x \to y^2 = Cx - x^2\to y =\sqrt{Cx - x^2}$$  use the initial condition $y= 4, x = 2$ implies $4 = \sqrt{2C - 4}$ gives you $C = 10.$ therefore the solution is $$ y=\sqrt{10x - x^2}$$
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