← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q6a — Step-by-Step Solution

15 marks · Section B

Quasilinear first-order PDEs (Lagrange's method) · PDEs · asked 9× in 13 yrs · Read the full method →

Question

Find the integral surface of the partial differential equation (x−y)y2∂z∂x+(y−x)x2∂z∂y=(x2+y2)z(x-y)y^2\frac{\partial z}{\partial x}+(y-x)x^2\frac{\partial z}{\partial y}=(x^2+y^2)z that contains the curve xz=a3xz=a^3, y=0y=0 on it.

Technique

Lagrange auxiliary equations; first integral x3+y3x^3+y^3 from cancelling (x−y)(x-y); second integral z/(x−y)z/(x-y) (found by multipliers / verified directly); eliminate parameters using the data curve.

Solution

This is a quasi-linear (Lagrange) equation Pp+Qq=RPp+Qq=R with

P=(x−y)y2,Q=(y−x)x2=−(x−y)x2,R=(x2+y2)z.P=(x-y)y^2,\qquad Q=(y-x)x^2=-(x-y)x^2,\qquad R=(x^2+y^2)z.

Step 1 — Lagrange’s auxiliary equations

dx(x−y)y2=dy−(x−y)x2=dz(x2+y2)z.\frac{dx}{(x-y)y^2}=\frac{dy}{-(x-y)x^2}=\frac{dz}{(x^2+y^2)z}.

Step 2 — First integral

From the first two ratios the common factor (x−y)(x-y) cancels:

dxy2=dy−x2 ⇒ x2 dx+y2 dy=0 ⇒ x33+y33=const.\frac{dx}{y^2}=\frac{dy}{-x^2}\ \Rightarrow\ x^2\,dx+y^2\,dy=0\ \Rightarrow\ \frac{x^3}{3}+\frac{y^3}{3}=\text{const}. u≡x3+y3=c1.\boxed{u\equiv x^3+y^3=c_1.}

Step 3 — Second integral

Use the multipliers (y,x,?)(y,x,?) on the first two members and pair with the third. Compute

y P+x Q=y(x−y)y2+x [−(x−y)x2]=(x−y)(y3−x3)=−(x−y)2(x2+xy+y2).y\,P+x\,Q=y(x-y)y^2+x\,[-(x-y)x^2]=(x-y)(y^3-x^3)=-(x-y)^2(x^2+xy+y^2).

With numerator y dx+x dy=d(xy)y\,dx+x\,dy=d(xy):

d(xy)−(x−y)2(x2+xy+y2)=dz(x2+y2)z.\frac{d(xy)}{-(x-y)^2(x^2+xy+y^2)}=\frac{dz}{(x^2+y^2)z}.

This still carries the factor (x−y)2(x-y)^2; a cleaner integral is found by testing zx−y\dfrac{z}{x-y}. Indeed, with ϕ=zx−y\phi=\dfrac{z}{x-y},

Pϕx+Qϕy+Rϕz=0(checked below),P\phi_x+Q\phi_y+R\phi_z=0\quad(\text{checked below}),

so

v≡zx−y=c2.\boxed{v\equiv\frac{z}{x-y}=c_2.}

(Verification: ϕx=−z(x−y)2\phi_x=\dfrac{-z}{(x-y)^2}, ϕy=z(x−y)2\phi_y=\dfrac{z}{(x-y)^2}, ϕz=1x−y\phi_z=\dfrac{1}{x-y}. Then Pϕx+Qϕy=−z(x−y)2[P−Q]=−z(x−y)2(x−y)(y2+x2)=−z(x2+y2)x−yP\phi_x+Q\phi_y=\dfrac{-z}{(x-y)^2}\big[P-Q\big]=\dfrac{-z}{(x-y)^2}(x-y)(y^2+x^2)=-\dfrac{z(x^2+y^2)}{x-y}, and Rϕz=(x2+y2)zx−yR\phi_z=\dfrac{(x^2+y^2)z}{x-y}; the sum is 00.) ✓

Step 4 — Apply the data curve y=0, xz=a3y=0,\ xz=a^3

On the curve y=0y=0, z=a3/xz=a^3/x. Evaluate the two integrals:

u=x3+0=x3,v=zx−0=zx=a3x2.u=x^3+0=x^3,\qquad v=\frac{z}{x-0}=\frac{z}{x}=\frac{a^3}{x^2}.

Eliminate xx: from u=x3u=x^3 we get x=u1/3x=u^{1/3}, so x2=u2/3x^2=u^{2/3} and

v=a3u2/3 ⇒ v u2/3=a3 ⇒ v3u2=a9.v=\frac{a^3}{u^{2/3}}\ \Rightarrow\ v\,u^{2/3}=a^3\ \Rightarrow\ v^3u^2=a^9.

Step 5 — Integral surface

Replace u=x3+y3u=x^3+y^3, v=z/(x−y)v=z/(x-y):

(zx−y)3(x3+y3)2=a9.\left(\frac{z}{x-y}\right)^3(x^3+y^3)^2=a^9.

Answer

  (x3+y3)2 z3=a9 (x−y)3.  \boxed{\;(x^3+y^3)^2\,z^3=a^9\,(x-y)^3.\;}
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