UPSC Maths 2025 Paper 2 Q7a — Solution

15 marks · Section B

Question

Find the complete integral of $z(p^2 - q^2) = x - y$ ; $p \equiv \dfrac{\partial z}{\partial x}$ , $q \equiv \dfrac{\partial z}{\partial y}$ .

Technique

Apply Charpit’s method for a nonlinear first-order PDE. The equation has a separable structure (each side splits into an $x,p,z$ part and a $y,q,z$ part), giving a first integral that leads directly to the complete integral.

Solution

Write $F(x,y,z,p,q) = z(p^2 - q^2) - (x - y) = zp^2 - zq^2 - x + y = 0.$

Charpit’s auxiliary equations

$\frac{dp}{F_x + pF_z} = \frac{dq}{F_y + qF_z} = \frac{dz}{-pF_p - qF_q} = \frac{dx}{-F_p} = \frac{dy}{-F_q}.$

The partials are $F_x = -1,\quad F_y = 1,\quad F_z = p^2 - q^2,\quad F_p = 2zp,\quad F_q = -2zq.$

Separable first integral

The PDE separates as $zp^2 - x = zq^2 - y \;(= a, \text{ a constant}),$ since the left side depends on $(x,z,p)$ and the middle on $(y,z,q)$ and they are equal. Then $zp^2 - x = a \implies p = \sqrt{\frac{x + a}{z}},\qquad zq^2 - y = a \implies q = \sqrt{\frac{y + a}{z}}.$

This is consistent: $z(p^2 - q^2) = (x + a) - (y + a) = x - y$ ✓, so the original PDE holds for every $a$ .

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

$dz = \sqrt{\frac{x + a}{z}}\,dx + \sqrt{\frac{y + a}{z}}\,dy \implies \sqrt{z}\,dz = \sqrt{x + a}\,dx + \sqrt{y + a}\,dy.$

Integrate each side: $\frac{2}{3}z^{3/2} = \frac{2}{3}(x + a)^{3/2} + \frac{2}{3}(y + a)^{3/2} + \frac{2}{3}b,$ where $b$ is the second arbitrary constant. Multiply by $3/2$ : $z^{3/2} = (x + a)^{3/2} + (y + a)^{3/2} + b.$

Complete integral

$\boxed{\,z^{3/2} = (x + a)^{3/2} + (y + a)^{3/2} + b\,}$ with $a, b$ arbitrary constants (a two-parameter family — the complete integral).

Answer

The complete integral is $z^{3/2} = (x + a)^{3/2} + (y + a)^{3/2} + b,\qquad a,b \text{ arbitrary constants}.$