Question

Solve the partial differentiatial equation
$( {x }^{2} + {y)}^{2} p + 2xyq = (x + y)z$
​

Answer 1

Step-by-step explanation:

Given

(x²+y²) p + 2xyq = (x+y)z

The subsidiary equations are

dx/(x²+y²) = dy/2xy = dz/(x+y)z

= [zdx + zdy - (x+y)dz]/[z(x²+y²)+2xyz- (x+y)²z]

= 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

dx/x²+y² = 2dy/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)

Which is required solution .