Answer to Question #143100 in Differential Equations for Nikhil

Question

Using the transformation T=z^2/2, reduce the equation f(x,y,z,p,q)= x(y^2+z^2×q^2)- zy^2×p=0 to a form f(P,x)= g(Q,y) where p= ∂T/∂x, Q= ∂T/∂y

Answer 1

Given equation is,

$f(x,y,z,p,q) = x(y^2+z^2×q^2)- zy^2p= xy^2+xz^2q^2- zy^2p$

Since, Transformation is, $T = \frac{z^2}{2}$

Now,

$z = \sqrt{2T}$

$p =\frac{\partial z}{\partial x} = \sqrt{2}\frac{1}{2} T^{-1/2}\frac{\partial T}{\partial x} =\frac{1}{\sqrt{2T}} \frac{\partial T}{\partial x} =\frac{1}{\sqrt{2T}} P$

$q=\frac{\partial z}{\partial y} = \sqrt{2}\frac{1}{2} T^{-1/2}\frac{\partial T}{\partial y} =\frac{1}{\sqrt{2T}} \frac{\partial T}{\partial y} =\frac{1}{\sqrt{2T}} Q$

Putting value of z, p and q in the given equation,

$xy^2+x(2T)(\frac{1}{\sqrt{2T}} Q)^2 -\sqrt{2T} y^2 (\frac{1}{\sqrt{2T}} P)=0$

$xy^2+x(2T)(\frac{1}{{2T}} Q^2) - y^2 ( P)=0$

$xy^2+xQ^2 - y^2 P=0$

$y^2(x-P) = -xQ^2$

$\frac{y^2}{Q^2} = \frac{x}{(P-x)}$

It can written as,

$g(Q,y) = f(P,x)$

which is the required result.