Non Linear PDE of first order

$$ \begin{align*} f(p,q)=0\\ \text{Assume } p=a\\ f(a,q)=0 \Rightarrow q=\phi(a)\\ \\ \delta z=\frac{\partial z}{\partial x}\delta x+\frac{\partial z}{\partial y}\delta y\\ \delta z =adx+\phi(a)dy\\ \text{integrating:}\\ z=ax+\phi(a)y+b \end{align*} $$

which is the complete solution since it takes the form $F(x,y,z,a,b)=0$ where a,b are arbitrary constants.

Various forms and how to deal with them #

f(z,p,q)=0 #

Assume $q=ap$

$$ \begin{align*} F(z,p,ap)=0\\ p=\phi(z)\\ \\ dz=pdx+qdy\\ dz=pdx+apdy\\ dz=\phi(z)(dx+ady)\\ dx+ady=\frac{dz}{\phi(z)}\\ \text{Integrating:}\\ x+ay=\int \frac{dz}{\phi(z)} +b \end{align*} $$

We obtain a complete solution.

Approach: One can solve it by considering a new function $u=x+ay$ to make $p=\frac{\partial z}{\partial u};q=a\frac{\partial z}{\partial u}$ and substituting it in f(z,p,q)=0.

f(x,p)=g(y,q)=a #

If we were able to separate x,p and y,q from each other $$ \begin{align*} f(x,p)=g(y,q)=a\\

f(x,p)=a\\ p=f_1(a,x)\\ \\ g(y,q)=a\\ q=g_1(a,y)\\ \\ dz=pdx+qdy\\ f_1(a,x)dx+g_1(a,y)dy\\ z=\int f_{1}(a,x)dx + \int g_{1}(a,y)dy + b \end{align*} $$

We obtain a complete solution.

Clairaut’s Form #

Similar to First order higher degree (nonlinear))#Clairaut’s Equation

z=px+qy+f(p,q) is Clairaut’s form

Complete solution: z=ax+by+f(a,b)