Question

Solve (3x+y-z)p + (x+y-z)q = 2(z+y)

Answer 1

$\frac{dx}{3x+yz}=\frac{dy}{x+yz}$

$(x+yz)dx=(3x+yz)dy$

$\frac{x^2}{2}+xyz-3xy-\frac{y^2z}{2}=C_1$

$\frac{dy}{x+yz}=\frac{dz}{2(z+y)}$

$(x+yz)dz=2(z+y)dy$

$xz+\frac{yz^2}{2}-2zy-y^2=C_2$

Answer:$\psi(\frac{x^2}{2}+xyz-3xy-\frac{y^2z}{2};xz+\frac{yz^2}{2}-2zy-y^2)$ =0