Answer to Question #150893 in Differential Equations for Ashweta Padhan

Question

Find the integral surface of the differential equation (x-y)p+(y-x-z)q=z passing through the circle z=1, x^2+y^2=1

Answer 1

To find the general solution of this differential equation and the particular solution through the aforementioned circle.

The corresponding symmetric system for the equation is

$$\frac{dx}{x - y}= \frac{dy}{(y - x - z) }= \frac{dz}{z}$$

and you need 2 independent first integrals.

Adding all numerators and denominators we get

$$\frac{(dx + dy + dz)}{(x - y + y - x - z + z)} = \frac{d(x + y + z)}{0}$$

what means

$d(x + y + z) = 0$ , so the $1^{st}$ independent integral is $f₁ = x + y + z.$

Next $\frac{(dx - dy + dz)}{(x - y - y + x + z + z)} = \frac{dz}{z}$ yields

$$\frac{d(x - y + z)}{(x - y + z)} = \frac{2 dz}{z}\ or\ d(\frac{ln|x - y + z|}{z²}) = 0$$

and we get the $2^{nd}$ independent integral $f₂ = \frac{x - y + z}{z²}$

According the theory of 1st order linear non-homogeneous partial differential equations the general solution in implicit form is $Φ((x - y + z)/z², x + y + z) = 0$ , here Φ is a differentiable function of 2 variables, or (x - y + z)/z² = φ(x + y + z), where φ is an arbitrary differentiable function of 1 variable.

Now we must determine φ to satisfy the initial condition. For z = 1 we get

x - y + 1 = φ(x + y + 1). If x + y + 1 = t and y = ±√(1 - x²), then

φ(t) = 1 ± √(1 + 2t - t²), finally

(x - y + z)/z² = 1 ± √(1 + 2(x+y+z) - (x+y+z)²) or

(x - y +z - z²)² = z⁴(1 + 2(x + y + z) - (x + y + z)²) in implicit form.