I am unable to solve this Lagrange PDE: $$ x^3 p + y(3 x^2 + y) q = z(2 x^2 + y), $$ where $p = \dfrac{\partial z}{\partial x}$, $q = \dfrac{\partial z}{\partial y}$.
I have found the $c_1 = \dfrac{xz}{y}$.
Unable to find the multipliers to find out $c_2$ to get the complete integral $f(c_1,c_2) = 0$.
$$x^3z_x+y(3x^2+y)z_y=z(2x^2+y)$$ The characteristic relationships are : $$\frac{dx}{x^3}=\frac{dy}{y(3x^2+y)}=\frac{dz}{z(2x^2+y)}$$ A first characteristic equation comes from $\frac{dx}{x^3}=\frac{dy}{y(3x^2+y)}$ Solving this first order linear ODE gives : $$x+\frac{x^3}{y}=c_1$$ A second characteristic equation is more difficult to derive. We need a combination of the above differential equations : $$\frac{ydx}{yx^3}=\frac{-xdy}{-xy(3x^2+y)}=\frac{ydx-xdy}{yx^3-xy(3x^2+y)}= \frac{ydx-xdy}{-xy(2x^2+y)}$$ $$\frac{ydx-xdy}{-xy(2x^2+y)}=\frac{dz}{z(2x^2+y)}$$ $$\frac{ydx-xdy}{-xy}=\frac{dz}{z}$$ $$\ln|z|+\ln|x|-\ln|y|=\text{constant}$$ $$z\frac{x}{y}=c_2$$ The general solution expressed on implicit form is : $$\Phi\left(x+\frac{x^3}{y}\:,\:z\frac{x}{y}\right)=0$$ where $\Phi$ is any differentiable function of two variables.
This implicit equation can be solved for $z$ , which gives the solution: $$z(x,y)=\frac{y}{x}F\left(x+\frac{x^3}{y}\right)$$ where $F$ is any differentiable function.
