Solve $\frac{dx}{x+z}=\frac{dy}{y}=\frac{dz}{y^2+z}$

Question

$$\frac{dx}{x+z}=\frac{dy}{y}=\frac{dz}{y^2+z}$$ Just a standard largarange auxillairy equation of the pde of quassi linear form, i tried grouping ,basic properties of ratios,but nothing seems to work.

Answer 1

Take the last two equations: $$ \frac{dy}{y} = \frac{dz}{y^2+z} $$ Rearrange the cross-ratio: $$ dy - \frac{ydz - zdy}{y^2} = 0 $$ The second term is the differential of the ratio, after which integration becomes straightforward: $$ dy - d\left(\frac{z}{y}\right) = 0 $$ $$ z = y(y-C_1) $$ Now take the first two equation: $$ \frac{dx}{x+z} = \frac{dy}{y} $$ $$ zdy = ydx - xdy $$ $$ \frac{z}{y}\frac{dy}{y} = \frac{ydx - xdy}{y^2} $$ Now using the first integral: $$ \frac{(y-C_1)dy}{y} = d\left(\frac{x}{y}\right) $$ Finally: $$ y-C_1\ln{|y|} = \frac{x}{y}+C_2 $$

Answer 2

$$\frac{dx}{x+z}=\frac{dy}{y}=\frac{dz}{y^2+z}$$ I suppose that the above Charpit-Lagrange system of ODEs comes from solving the PDE : $$(x+z)\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=y^2+z \tag 1$$ A first characteristic equation comes from $\frac{dy}{y}=\frac{dz}{y^2+z}$

$\frac{dz}{dy}=y-\frac{z}{y}\quad$ is a first order linear ODE easy to solve : $z=y^2+c_1y$ $$\frac{z}{y}-y=c_1$$

A second characteristic equation comes from $\frac{dx}{x+z}=\frac{dy}{y}=\frac{dx}{x+(y^2+c_1y)}$

$\frac{dx}{dy}=\frac{x}{y}+y+c_1$ is a first order linear ODE easy to solve for $x(y)$ :

$\frac{x}{y}-y-c_1\ln|y|=c_2$ $$\frac{x}{y}-y-(\frac{z}{y}-y)\ln|y|=c_2$$ The general solution of the PDE $(1)$ on the form of implicit equation $\Phi(c_1,c_2)=0$ is : $$\Phi\left((\frac{z}{y}-y) \:,\:(\frac{x}{y}-y-(\frac{z}{y}-y)\ln|y|)\right)=0$$ $\Phi$ is an arbitrary function of two variables.

$\Phi$ has to be determined according to some boundary condition.