Answer to Question #164224 in Differential Equations for hasham

Question

Obtain PDE z = x + ax² y² where a and b are arbitrary constants and also

analyze the partial differential equation. 

Answer 1

Given equation is-

$z=x+ax^2y^2$

Differentiate w.r.t. to x partially

$\dfrac{dz}{dx}=1+2axy^2$

$(\dfrac{dz}{dx}-1)=2axy^2$

$a=\dfrac{1}{2xy^2}(\dfrac{dz}{dx}-1)$

Putting the value of a in given equation-

$z=x+\dfrac{1}{2xy^2}(p-1)x^2y^2$ {$\dfrac{dz}{dx}=p$ }

=$x+\dfrac{1}{2}(p-1)x$

$=x+\dfrac{px}{2}-\dfrac{x}{2}$

z= $\dfrac{x(1+p)}{2}$