Question

(px-y)( x-py)=2p
Using x^2=u,y^2=u
Find general solution and solveble solutin

Answer 1

$$(px-y)(x-py)=2p$$

$$(pxy-y^2)(x-py)=2py$$

$$({py \over x}\cdot x^2-y^2)(1-{py \over x})=2{py \over x}$$

$${py \over x}\cdot x^2-y^2=\dfrac{2\dfrac{py}{x}}{1-\dfrac{py}{x}}$$

$$y^2={py \over x}\cdot x^2-\dfrac{2\dfrac{py}{x}}{1-\dfrac{py}{x}}$$

Let $x^2=u, y^2=v.$ Then

$$du=2xdx, dv=2ydy$$

$$p=\dfrac{dy}{dx}=\dfrac{2x}{2y}\cdot \dfrac{dv}{du}=\dfrac{\sqrt{u}}{\sqrt{v}}\cdot P, P=\dfrac{dv}{du}$$

$${py \over x}=\dfrac{dv}{du}=P$$

Substitute

$$v=Pu-\dfrac{2P}{1-P}$$

We have Clairaut Equation

$$v=Pu+f(P)$$

The general solution is given by $v=Cu+f(C), C$ is an arbitrary constant.

Therefore

$$v=Cu-\dfrac{2C}{1-C}, C \ \text{is arbitrary constant}, C\not=1$$

The general solution is

$$y^2=Cx^2-\dfrac{2C}{1-C}, \ C \ \text{is arbitrary constant}, C\not=1$$