← 2017 Paper 2
UPSC 2017 Maths Optional Paper 2 Q6a — Step-by-Step Solution 15 marks · Section B
Charpit's method · PDEs · asked 2× in 14 yrs · Read the full method →
Question
Find a complete integral of the partial differential equation 2 ( p q + y p + q x ) + x 2 + y 2 = 0 2(pq+yp+qx)+x^2+y^2=0 2 ( pq + y p + q x ) + x 2 + y 2 = 0
Technique
Charpit’s method; the auxiliary system yields the clean first integral ( p + x ) + ( q + y ) = a (p+x)+(q+y)=a ( p + x ) + ( q + y ) = a F = 2 ( p + x ) ( q + y ) + ( x − y ) 2 F=2(p+x)(q+y)+(x-y)^2 F = 2 ( p + x ) ( q + y ) + ( x − y ) 2 p , q p,q p , q d z = p d x + q d y dz=p\,dx+q\,dy d z = p d x + q d y sinh − 1 \sinh^{-1} sinh − 1
Solution
Setup. Write the equation as
F ( x , y , z , p , q ) = 2 p q + 2 y p + 2 q x + x 2 + y 2 = 0 , p = ∂ z ∂ x , q = ∂ z ∂ y . F(x,y,z,p,q)=2pq+2yp+2qx+x^2+y^2=0,\qquad p=\frac{\partial z}{\partial x},\ q=\frac{\partial z}{\partial y}. F ( x , y , z , p , q ) = 2 pq + 2 y p + 2 q x + x 2 + y 2 = 0 , p = ∂ x ∂ z , q = ∂ y ∂ z . F F F z z z F z = 0 F_z=0 F z = 0
Step 1 — Charpit’s auxiliary equations
Charpit’s subsidiary system is
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 . Compute the partials:
F x = 2 q + 2 x , F y = 2 p + 2 y , F p = 2 q + 2 y , F q = 2 p + 2 x , F z = 0. F_x=2q+2x,\quad F_y=2p+2y,\quad F_p=2q+2y,\quad F_q=2p+2x,\quad F_z=0. F x = 2 q + 2 x , F y = 2 p + 2 y , F p = 2 q + 2 y , F q = 2 p + 2 x , F z = 0. So
d p 2 ( q + x ) = d q 2 ( p + y ) = d x − 2 ( q + y ) = d y − 2 ( p + x ) . \frac{dp}{2(q+x)}=\frac{dq}{2(p+y)}=\frac{dx}{-2(q+y)}=\frac{dy}{-2(p+x)}. 2 ( q + x ) d p = 2 ( p + y ) d q = − 2 ( q + y ) d x = − 2 ( p + x ) d y . Step 2 — A simple first integral
Add the p p p x x x q q q y y y = 1 2 d t =\tfrac12\,dt = 2 1 d t
d p = 2 ( q + x ) d t , d x = − 2 ( q + y ) d t ⇒ d ( p + x ) = 2 ( x − y ) d t , dp=2(q+x)\,dt,\quad dx=-2(q+y)\,dt\ \Rightarrow\ d(p+x)=2(x-y)\,dt, d p = 2 ( q + x ) d t , d x = − 2 ( q + y ) d t ⇒ d ( p + x ) = 2 ( x − y ) d t , d q = 2 ( p + y ) d t , d y = − 2 ( p + x ) d t ⇒ d ( q + y ) = 2 ( y − x ) d t . dq=2(p+y)\,dt,\quad dy=-2(p+x)\,dt\ \Rightarrow\ d(q+y)=2(y-x)\,dt. d q = 2 ( p + y ) d t , d y = − 2 ( p + x ) d t ⇒ d ( q + y ) = 2 ( y − x ) d t . Hence d ( p + x ) + d ( q + y ) = 0 d(p+x)+d(q+y)=0 d ( p + x ) + d ( q + y ) = 0
( p + x ) + ( q + y ) = a ( constant ) . (p+x)+(q+y)=a\quad(\text{constant}). ( p + x ) + ( q + y ) = a ( constant ) . Step 3 — Use the PDE to factorise
Set P = p + x , Q = q + y P=p+x,\ Q=q+y P = p + x , Q = q + y
2 P Q = 2 ( p + x ) ( q + y ) = 2 p q + 2 p y + 2 x q + 2 x y , 2PQ=2(p+x)(q+y)=2pq+2py+2xq+2xy, 2 P Q = 2 ( p + x ) ( q + y ) = 2 pq + 2 p y + 2 x q + 2 x y , so
F = 2 p q + 2 y p + 2 q x + x 2 + y 2 = 2 P Q − 2 x y + x 2 + y 2 = 2 P Q + ( x − y ) 2 . F=2pq+2yp+2qx+x^2+y^2=2PQ-2xy+x^2+y^2=2PQ+(x-y)^2. F = 2 pq + 2 y p + 2 q x + x 2 + y 2 = 2 P Q − 2 x y + x 2 + y 2 = 2 P Q + ( x − y ) 2 . Therefore the PDE F = 0 F=0 F = 0
2 ( p + x ) ( q + y ) + ( x − y ) 2 = 0 ⟹ P Q = − 1 2 ( x − y ) 2 . 2(p+x)(q+y)+(x-y)^2=0\ \Longrightarrow\ PQ=-\tfrac12(x-y)^2. 2 ( p + x ) ( q + y ) + ( x − y ) 2 = 0 ⟹ P Q = − 2 1 ( x − y ) 2 . Step 4 — Solve for P , Q P,Q P , Q p , q p,q p , q
With P + Q = a P+Q=a P + Q = a P Q = − 1 2 ( x − y ) 2 PQ=-\tfrac12(x-y)^2 P Q = − 2 1 ( x − y ) 2 P , Q P,Q P , Q t 2 − a t − 1 2 ( x − y ) 2 = 0 t^2-at-\tfrac12(x-y)^2=0 t 2 − a t − 2 1 ( x − y ) 2 = 0
P = a + a 2 + 2 ( x − y ) 2 2 , Q = a − a 2 + 2 ( x − y ) 2 2 . P=\frac{a+\sqrt{a^2+2(x-y)^2}}{2},\qquad Q=\frac{a-\sqrt{a^2+2(x-y)^2}}{2}. P = 2 a + a 2 + 2 ( x − y ) 2 , Q = 2 a − a 2 + 2 ( x − y ) 2 . Thus
p = P − x , q = Q − y . p=P-x,\qquad q=Q-y. p = P − x , q = Q − y . Step 5 — Integrate d z = p d x + q d y dz=p\,dx+q\,dy d z = p d x + q d y
d z = ( P − x ) d x + ( Q − y ) d y = a 2 ( d x + d y ) − ( x d x + y d y ) + 1 2 a 2 + 2 ( x − y ) 2 ( d x − d y ) . dz=(P-x)\,dx+(Q-y)\,dy=\frac a2(dx+dy)-(x\,dx+y\,dy)+\frac12\sqrt{a^2+2(x-y)^2}\,(dx-dy). d z = ( P − x ) d x + ( Q − y ) d y = 2 a ( d x + d y ) − ( x d x + y d y ) + 2 1 a 2 + 2 ( x − y ) 2 ( d x − d y ) . Let s = x − y s=x-y s = x − y d x − d y = d s dx-dy=ds d x − d y = d s s s s
z = a 2 ( x + y ) − x 2 + y 2 2 + 1 2 ∫ a 2 + 2 s 2 d s + b , z=\frac a2(x+y)-\frac{x^2+y^2}{2}+\frac12\int\sqrt{a^2+2s^2}\,ds+b, z = 2 a ( x + y ) − 2 x 2 + y 2 + 2 1 ∫ a 2 + 2 s 2 d s + b , with
∫ a 2 + 2 s 2 d s = s 2 a 2 + 2 s 2 + a 2 2 2 sinh − 1 2 s a . \int\sqrt{a^2+2s^2}\,ds=\frac{s}{2}\sqrt{a^2+2s^2}+\frac{a^2}{2\sqrt2}\sinh^{-1}\!\frac{\sqrt2\,s}{a}. ∫ a 2 + 2 s 2 d s = 2 s a 2 + 2 s 2 + 2 2 a 2 sinh − 1 a 2 s . Step 6 — Complete integral
Answer
z = a 2 ( x + y ) − x 2 + y 2 2 + x − y 4 a 2 + 2 ( x − y ) 2 + a 2 4 2 sinh − 1 2 ( x − y ) a + b , \boxed{\;z=\frac a2(x+y)-\frac{x^2+y^2}{2}+\frac{x-y}{4}\sqrt{a^2+2(x-y)^2}+\frac{a^2}{4\sqrt2}\sinh^{-1}\!\frac{\sqrt2\,(x-y)}{a}+b,\;} z = 2 a ( x + y ) − 2 x 2 + y 2 + 4 x − y a 2 + 2 ( x − y ) 2 + 4 2 a 2 sinh − 1 a 2 ( x − y ) + b ,