Question

(4z + x²z) dz + (1+z²)dx - (yz² + y)(4+x²)dy=0

Answer 1

By Lagrange method,

$\frac{dx}{1+z^2}=\frac{dy}{- (yz² + y)(4+x²)}=\frac{dz}{4z+x^2z}\\$

By taking the first equation,

$\frac{dy}{- (yz² + y)(4+x²)}=\frac{dz}{4z+x^2z}\\ \frac{dy}{- y(z² + 1)(4+x²)}=\frac{dz}{z(4+x^2)}\\ \frac{dy}{- y}=(\frac{1}{z}+z)dz\\$

Integrating both sides,

$-lny=lnz+\frac{z^2}{2}+c\\ ln(\frac{1}{yz})-\frac{z^2}{2}=c$

And, from the second equation,

$\frac{dx}{1+z^2}=\frac{dz}{4z+x^2z}\\ \frac{dx}{1+z^2}=\frac{dz}{z(4+x^2)}\\ (4+x^2)dx=(\frac{1}{z}+z)dz\\ 4x+\frac{x^3}{3}=lnz+\frac{z^2}{2}+c\\ 4x+\frac{x^3}{3}-lnz-\frac{z^2}{2}=c$

Thus, the general solution is given by

$F(ln(\frac{1}{yz})-\frac{z^2}{2}, 4x+\frac{x^3}{3}-lnz-\frac{z^2}{2})=0$