Question

(x^2+y^2)p +2xyq=(x+y)z

Answer 1

The given equation is

     $(x²+y²) p + 2xyq = (x+y)z$

 The subsidiary equations are 

 $\dfrac{dx}{x²+y²} = \dfrac{dy}{2xy} = \dfrac{dz}{(x+y)z}$

   = $\dfrac{[zdx + zdy - (x+y)dz]}{[z(x²+y²)+2xyz- (x+y)²z]}$

   = $\dfrac{z(dx+dy) - (x+y)dz}{0}$

⇒z((dx+dy) - (x+y) )dz = 0

⇒z d(x+y) - (x+y)dz = 0

⇒(zd(x+y) - (x+y)dz) / z² = 0

⇒d((x+y) / z) = 0 

⇒(x+y) / z = c₁ 

From first two ratios we get 

$\dfrac{dx}{x²+y²} = 2\dfrac{dy}{4xy}$

 ⇒2(x²+y²) dy = 4xy dx

 ⇒2(x²+y²) dy - 4xy dx = 0

 ⇒2x² dy - 2y² dy + 4y² dy -4xy dx = 0

 ⇒2(x²-y²) dy - 2y(2xdx - 2ydy) = 0

 ⇒[(x²-y²) 2 dy - 2y d(x²-y²)]/(x²-y²)²

                 = 0/(x²-y²)

 ⇒d(2y / (x²-y²)) = 0

Integrating , we get 

  2y/(x²-y²) = c₂ 

∴ The solution is (x+y) / z = Φ(2y / (x²-y²))