# How to solve this PDE: $p\sqrt x+q\sqrt y =\sqrt z$

I am trying to solve this PDE, using Lagrange's method: $p\sqrt x+q\sqrt y =\sqrt z$

Lagrange's auxiliary equation is: $\frac{dx}{\sqrt x}=\frac {dy}{\sqrt y}= \frac{dz}{\sqrt z}$

I already know the answer to this on which is $f(\sqrt x - \sqrt y, \sqrt y - \sqrt z)$ , but I don't know the complete solution and I have tried to solve it but I can't quite get to the same answer.

• What is it PDE? – Michael Rozenberg Jan 15 '18 at 17:45
• What you've posted does not contain any partial derivatives... Try reading the question again? – AlkaKadri Jan 15 '18 at 17:46
• This is a partial differential equation of the form Pp+Qq=R, where P, Q and R are functions of x, y, z. – House A. Jan 15 '18 at 17:51
• What are $p$ and $q$? – Bernard Massé Jan 15 '18 at 17:53
• @MichaelRozenberg Partial Differential Equation. This is a standard form of a linear partial differential equation of the first order ( Lagrange's linear equation ). – House A. Jan 15 '18 at 17:55

$$\sqrt{x} \frac{\partial z}{\partial x}+\sqrt{y} \frac{\partial z}{\partial y} =\sqrt z$$ Your general solution on the form of implicit equation : $\quad f(\sqrt x - \sqrt y, \sqrt y - \sqrt z)=0\quad$ is correct.

An equivalent explicit form is : $$\sqrt y - \sqrt z =\phi( \sqrt x - \sqrt y)$$ where $\phi$ is any differentiable function.

As a consequence : $$z(x,y)=\bigg(\sqrt y - \phi( \sqrt x - \sqrt y)\bigg)^2$$

System of characteristic ODEs : $\quad\frac{dx}{\sqrt x}=\frac{dy}{\sqrt y}=\frac{dz}{\sqrt z}$
First family of characteristic curves, from $\quad \frac{dx}{\sqrt x}=\frac{dy}{\sqrt y}\quad\to\quad \sqrt x-\sqrt y=c_1$
Second family of characteristic curves, from $\quad \frac{dz}{\sqrt z}=\frac{dy}{\sqrt y}\quad\to\quad \sqrt z-\sqrt y=c_2$
General solution : $\quad\sqrt y - \sqrt z =\phi( \sqrt x - \sqrt y)$
$$z(x,y)=\bigg(\sqrt y - \phi( \sqrt x - \sqrt y)\bigg)^2$$