Answer to Question #106269 in Differential Equations for Suraj Singh

Question

Solve: z(p-q) = z^2 + (x+(y^2))

Answer 1

$pz-qz=z^2+(x+y)^2\\ \frac{dx}{z}=\frac{dy}{-z}=\frac{dz}{z^2+(x+y)^2}\\ 1) \frac{dx}{z}=\frac{dy}{-z}\implies\\ -dx=dy\implies -\int dx=\int dy\implies\\ -x+c_1=y \implies c_1=x+y\\ 2) \frac{dx}{z}=\frac{dz}{z^2+(x+y)^2} \implies\\ \frac{dx}{z}=\frac{dz}{z^2+c_1^2}\implies\\ dx=\frac{zdz}{z^2+c_1^2} \implies\\ \int dx=\int\frac{zdz}{z^2+c_1^2}\implies\\ \int dx=\frac{1}{2}\int\frac{d(z^2+c_1^2)}{z^2+c_1^2}\implies\\ \frac{1}{2}\ln|z^2+c_1^2|=x+\frac{1}{2}\ln|c_2|\implies\\ \ln|z^2+c_1^2|=2x+\ln|c_2|\implies\\ z^2+c_1^2=c_2e^{2x}\implies\\ z^2+(x+y)^2=c_2e^{2x}\implies\\ c_2=\frac{z^2+(x+y)^2}{e^{2x}}\\$

Solution of equation is

$F(x+y, \frac{z^2+(x+y)^2}{e^{2x}})=0$