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UPSC 2025 Maths Optional Paper 2 Q7a — Step-by-Step Solution 15 marks · Section B
Charpit's method · PDEs · asked 2× in 13 yrs · Read the full method →
Question
Find the complete integral of z ( p 2 − q 2 ) = x − y z(p^2 - q^2) = x - y z ( p 2 − q 2 ) = x − y p ≡ ∂ z ∂ x p \equiv \dfrac{\partial z}{\partial x} p ≡ ∂ x ∂ z q ≡ ∂ z ∂ y q \equiv \dfrac{\partial z}{\partial y} q ≡ ∂ y ∂ z
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 x,p,z x , p , z y , q , z y,q,z y , q , z
Solution
Write F ( x , y , z , p , q ) = z ( p 2 − q 2 ) − ( x − y ) = z p 2 − z q 2 − x + y = 0. F(x,y,z,p,q) = z(p^2 - q^2) - (x - y) = zp^2 - zq^2 - x + y = 0. F ( x , y , z , p , q ) = z ( p 2 − q 2 ) − ( x − y ) = z p 2 − z q 2 − x + y = 0.
Charpit’s auxiliary equations
d p F x + p F z = d q F y + q F z = d z − p F p − q F q = d x − F p = d y − F q . \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}. F x + p F z d p = F y + q F z d q = − p F p − q F q d z = − F p d x = − F q d y . The partials are
F x = − 1 , F y = 1 , F z = p 2 − q 2 , F p = 2 z p , F q = − 2 z q . F_x = -1,\quad F_y = 1,\quad F_z = p^2 - q^2,\quad F_p = 2zp,\quad F_q = -2zq. F x = − 1 , F y = 1 , F z = p 2 − q 2 , F p = 2 z p , F q = − 2 z q . Separable first integral
The PDE separates as
z p 2 − x = z q 2 − y ( = a , a constant ) , zp^2 - x = zq^2 - y \;(= a, \text{ a constant}), z p 2 − x = z q 2 − y ( = a , a constant ) , since the left side depends on ( x , z , p ) (x,z,p) ( x , z , p ) ( y , z , q ) (y,z,q) ( y , z , q )
z p 2 − x = a ⟹ p = x + a z , z q 2 − y = a ⟹ q = y + a z . zp^2 - x = a \implies p = \sqrt{\frac{x + a}{z}},\qquad zq^2 - y = a \implies q = \sqrt{\frac{y + a}{z}}. z p 2 − x = a ⟹ p = z x + a , z q 2 − y = a ⟹ q = z y + a . This is consistent: z ( p 2 − q 2 ) = ( x + a ) − ( y + a ) = x − y z(p^2 - q^2) = (x + a) - (y + a) = x - y z ( p 2 − q 2 ) = ( x + a ) − ( y + a ) = x − y a a a
Integrate d z = p d x + q d y dz = p\,dx + q\,dy d z = p d x + q d y
d z = x + a z d x + y + a z d y ⟹ z d z = x + a d x + y + a d y . 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. d z = z x + a d x + z y + a d y ⟹ z d z = x + a d x + y + a d y . Integrate each side:
2 3 z 3 / 2 = 2 3 ( x + a ) 3 / 2 + 2 3 ( y + a ) 3 / 2 + 2 3 b , \frac{2}{3}z^{3/2} = \frac{2}{3}(x + a)^{3/2} + \frac{2}{3}(y + a)^{3/2} + \frac{2}{3}b, 3 2 z 3/2 = 3 2 ( x + a ) 3/2 + 3 2 ( y + a ) 3/2 + 3 2 b , where b b b 3 / 2 3/2 3/2
z 3 / 2 = ( x + a ) 3 / 2 + ( y + a ) 3 / 2 + b . z^{3/2} = (x + a)^{3/2} + (y + a)^{3/2} + b. z 3/2 = ( x + a ) 3/2 + ( y + a ) 3/2 + b . Complete integral
z 3 / 2 = ( x + a ) 3 / 2 + ( y + a ) 3 / 2 + b \boxed{\,z^{3/2} = (x + a)^{3/2} + (y + a)^{3/2} + b\,} z 3/2 = ( x + a ) 3/2 + ( y + a ) 3/2 + b with a , b a, b a , b
Answer
The complete integral is
z 3 / 2 = ( x + a ) 3 / 2 + ( y + a ) 3 / 2 + b , a , b arbitrary constants . z^{3/2} = (x + a)^{3/2} + (y + a)^{3/2} + b,\qquad a,b \text{ arbitrary constants}. z 3/2 = ( x + a ) 3/2 + ( y + a ) 3/2 + b , a , b arbitrary constants .