Answer to Question #221431 in Differential Equations for Nikith

Question

Solve the PDE p+q=1

Answer 1

Given Partial Differential equation is -

$=p+q=1$

Let $p=\dfrac{\partial u}{\partial t}$ $, q=\dfrac{\partial u}{\partial x}$

$=$ $\dfrac{\partial u}{\partial t}+\dfrac{\partial u}{\partial x}=1$

This gives the system of ODE's as -

$=\dfrac{dx}{1}=\dfrac{dt}{1}=\dfrac{du}{1}$

which has corresponding equation equal to some constants from there integrals-

$=x-t=c_1$

$=u-x=c_2$

Which gives ,

$u(x,t)=x+{\phi}(x-t)$ , which is the required solution, where $\phi$ is an arbitrary function.