Answer to Question #146590 in Differential Equations for Ashweta Padhan

Question

p^2q^2+x^2y^2= x^2q^2(x^2+y^2)

Answer 1

$$Solution$$

To find the complete integral of the given PDE

The given PDE is

$$p^2q^2+x^2y^2= x^2q^2(x^2+y^2)$$

we may re-write it as

$$p^2q^2+x^2y^2=x^2q^2(x^2+y^2)\\ or\\ p^2q^2+x^2y^2=x^4q^2+x^2y^2q^2\\ or\\ p^2q^2=x^4q^2+x^2y^2(q^2-1)\\ or\\ p^2=x^4+x^2y^2(1-\frac{1}{q^2})\\ or\\ p^2-x^4=x^2y^2(1-\frac{1}{q^2})\\ or\\ \frac{p^2}{x^2}-x^2=y^2(1-\frac{1}{q^2})\\ or\\ \frac{p^2}{x^2}-x^2=y^2-\frac{y^2}{q^2}=a^2,\ say.$$

This is the special form $(f(p,x) =g(q,y)type)$ , of the Charpit’s equations,

$$\frac{p^2}{x^2}-x^2=a^2\ and\ y^2-\frac{y^2}{q^2}=a^2\\i.e\\ p=x\sqrt{x^2+a^2}\ and\ q=\frac{y}{\sqrt{y^2-a^2}}$$

Putting these values of $p$ and $q$ in the equation

$$dz=p\ dx+q\ dy$$

We have

$$dz=x\sqrt{x^2+a^2}\ dx+\frac{y}{\sqrt{y^2-a^2}}dy$$

Integrating it, we get the complete integral as

$$z=\frac13(x^2+a^2)^\frac32+\sqrt{y^2-a^2}+b$$

Is the complete integral f the PDE