Answer to Question #105029 in Differential Equations for Vaishali

Question

Find the integral surface of the partial differential equation
(x - y) y^2 p + (y - x) x^2 q = (x^2 + y^2) z
Through the curve xz = a^2 , y = 0

Answer 1

dx/((x-y)*y²)=dy/((y-x)*x²)=dz/((x²+y²)*z)

$\int$ y²dy = $\int$-x²dx

y³=-x³+c₁ (c₁ - constant) (1)

$\int$ y²/z*dz = $\int$ (x²+y²)/(x-y)dx

y²ln(z)=1/2*(x²-3y²+2xy+4y²ln(x-y))+c₂ (c₂ - constant) (2)

y³+x³ = c₁

y²ln(z) - 1/2*(x²-3y²+2xy+4y²ln(x-y)) = c₂

y=0

z = a²/x

c₁ = x³

x = c₁^1/3

c₂ = - 1/2*(x²)

x = (-2*c₂)^1/2

c₁^1/3 = (-2*c₂)^1/2

c₁² = (-2*c₂)³

x⁶=(-2*(y²ln(z) - 1/2*(x²-3y²+2xy+4y²ln(x-y))))³