Question

x² * p² - 2xyp + 2y² - x² = 0

Answer 1

Given equation is :

⟹ x²p² - 2xyp + 2y² - x² = 0

⟹ on solving we get p = ( 2xy + 2x(x² - y²)^1/2 )/2x²

⟹ dy/dx = (y ± (x² - y²)^1/2)/x ..........( 1)

which is homogeneous in x and y

put y=vx so that dy/dx= v+x*dv/dx

∴ from (1), v+dv/dx = (vx± (x²-v²x²)^1/2)/x = v ± (1- v)^1/2

⟹ x*dv/dx = ±(1 - v²)^1/2

⟹ dv/(1 - v²)^1/2 = ±dx/x

on integrating we get,

sin^-1 v = log x + log c or sin^-1 v= -log x - log c

sin^-1 y/x = ± log (cx) is the required solution