Answer to Question #220662 in Differential Equations for Ashish

Question

Equation 𝑓(𝑧, 𝑝, 𝑞) = 0 , it’s solution is given by

Answer 1

$$f(z, p, q)=0$$

Let $z=g(x+ay)$ be a trial solution of the equation $f(z, p, q)=0.$  

We, now put $x+ay=u,$ so that, we have

$$z=g(u), \dfrac{\partial u}{\partial x}=1, \dfrac{\partial u}{\partial y}=a$$

$$p=\dfrac{\partial z}{\partial x}=\dfrac{dz}{du}\cdot\dfrac{\partial u}{\partial x}=\dfrac{dz}{du}$$

$$q=\dfrac{\partial z}{\partial y}=\dfrac{dz}{du}\cdot\dfrac{\partial u}{\partial y}=a\dfrac{dz}{du}$$

Substituting the values of $p$ and $q$ in $f(z, p, q)=0,$ we get

$$f(z, \dfrac{dz}{du}, a\dfrac{dz}{du})=0$$

which is an ordinary differential equation of order one.

The solution of the equation

$$f(z, \dfrac{dz}{du}, a\dfrac{dz}{du})=0$$

is given by $z=g(u+b)$ or $z=g(ax+y+b),$ which is the complete integral of partial differential equation

$$f(z, p, q)=0$$