Question

Find the general integral of

2x(y+z²)p+y(2y+z²)q=z²

Answer 1

$2x(y+z^2)p+y(2y+z^2)q=z^2\\ 2x(y+z^2)\frac{\partial z}{\partial x}+y(2y+z^2)\frac{\partial z}{\partial y}=z^2\\ \frac{dx}{2x(y+z^2)}=\frac{dy}{y(2y+z^2)}=\frac{dz}{z^2}\\ \text{Multiply through by } \frac{xy}{xy}\\ y\frac{dx}{2xy^2+2xyz^2}=x\frac{dy}{2xy^2+xyz^2}=xy\frac{dz}{xyz^2}\\ \frac{ydx-xdy-xydz}{2xy^2+2xyz^2-2xy^2-xyz^2-xyz^2}=\frac{ydx-xdy-xydz}{0}..........(1)\\ ydx-xdy-xydz=0\\ \text{Divide through by } xy\\ \frac{dx}{x}-\frac{dy}{y}-dz=0\\ \text{Integrate through}\\ \ln x-\ln y-z=a\\ \ln (\frac{x}{y})-z=a$

Using $x, y, z$ as multipliers. Each fraction if (1) will be

$\frac{xydx-xydy-xyzdz}{2xy^2+2xyz^2-2xy^2-xyz^2-xyz^2}=\frac{xydx-xydy-xyzdz}{0}\\ xydx-xydy-xyzdz=0\\ \text{Divude through by } xy\\ dx-dy-zdz=0\\ \text{Integrate}\\ x-y-\frac{z^2}{2}=b$

Therefore, the general solution is $f(\ln (\frac{x}{y})-z, x-y-\frac{z^2}{2})=0$