Answer to Question #132479 in Differential Equations for Nikhil

Question

Find the complete integral of (x+y)(p+q)^2+(x-y)(p-q)^2=1

Answer 1

To solve differential equation $(x+y)(p+q)^2+(x-y)(p-q)^2 = 1$

Let us put $x+y = X^2$ and $x-y = Y^2$

then

$p = \frac{\partial z}{\partial x} = \frac{\partial z}{\partial X}\frac{\partial X}{\partial x}+ \frac{\partial z}{\partial Y}\frac{\partial Y}{\partial x} = \frac{1}{2X}\frac{\partial z}{\partial X}+ \frac{1}{2Y}\frac{\partial z}{\partial Y}$

$q = \frac{\partial z}{\partial y} = \frac{\partial z}{\partial X}\frac{\partial X}{\partial y}+ \frac{\partial z}{\partial Y}\frac{\partial Y}{\partial y} = \frac{1}{2X}\frac{\partial z}{\partial X}- \frac{1}{2Y}\frac{\partial z}{\partial Y}$

Then

$p+q = \frac{1}{X}\frac{\partial z}{\partial X}$

and

$p-q = \frac{1}{Y}\frac{\partial z}{\partial Y}$

Now substitute it in the equation, then we get

$(\frac{\partial z}{\partial X } )^2 + (\frac{\partial z}{\partial Y } )^2 = 1$

Complete integral of the equation is

$z = aX+bY+c$ where $a^2 + b^2 = 1$

Put value of X and Y, we get

$z = a\sqrt{(x+y)} + \sqrt{1-a^2}\sqrt{(x-y)} + c$

where a and c are arbitrary constants.