Question

Solve:(p − q)z = z^2 + (x + y)^2

Answer 1

Given equation is $(p-q)z = z^2 + (x+y)^2$

Comparing above equation with $Pp + Qq = R$

Then

$\frac{dx}{z}=\frac{dy}{-z}=\frac{dz}{z^2 + (x+y)^2}$

Taking first two,

$\frac{dx}{z} = - \frac{dy}{z}$

Integrating both sides,

$x = - y + C$

$x+y = C$ (1)

Then taking first and last,

$\frac{dx}{z} = \frac{dz}{z^2+(x+y)^2}$

$\frac{dx}{z} = \frac{dz}{z^2+C^2}$

Integrating both sides

$\int {dx}=\int \frac{zdz}{z^2+C^2}$

$x = \frac{1}{2}log|z^2 + C^2| + log|b|$

then $b = (z^2 + (x+y)^2)e^{-2x}$

Hence,

solution of the equation will be

$f(x+y) = (z^2 + (x+y)^2)e^{-2x}$