Question

Show that p=q(x+y/xy) is a sol'n to the partial diff. equation
x^2 dp/dx - y^2 dp/dy = 0.

This isn't true... right?

Answer 1

(x+y)/(xy)...?

Answer 2

when i calculated them, i got 2x^3y-2xy^3 = 0

Answer 3

= x/xy + y/xy = 1/y + 1/x

q*d/dx { ((1/y+1/x) }= -q/x^2

q*d/dy { ((1/y+1/x) }= -q/y^2

x^2 (-q/x^2) - y^2 (-q/y^2) = ?

Answer 4

hmmm ok.