← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

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

Question

Solve the partial differential equation

(y2+z2−x2) p−2xy q+2xz=0,p=∂z∂x,  q=∂z∂y.(y^2+z^2-x^2)\,p-2xy\,q+2xz=0,\quad p=\dfrac{\partial z}{\partial x},\;q=\dfrac{\partial z}{\partial y}.

Technique

Lagrange’s auxiliary system dx/P=dy/Q=dz/Rdx/P=dy/Q=dz/R; first integral from the cleanest ratio (dy/y=dz/zdy/y=dz/z); second integral via multipliers (x,y,z)(x,y,z) to spot d(x2+y2+z2)d(x^2+y^2+z^2).

Solution

Setup. Quasilinear first-order PDE in form Pp+Qq=RPp+Qq=R with P=y2+z2−x2P=y^2+z^2-x^2, Q=−2xyQ=-2xy, R=−2xzR=-2xz.

(Rewriting the original equation (y2+z2−x2)p−2xyq+2xz=0(y^2+z^2-x^2)p-2xy q+2xz=0 as (y2+z2−x2)p+(−2xy)q=−2xz(y^2+z^2-x^2)p+(-2xy)q=-2xz.)

Lagrange’s auxiliary system:

dxy2+z2−x2=dy−2xy=dz−2xz.\dfrac{dx}{y^2+z^2-x^2}=\dfrac{dy}{-2xy}=\dfrac{dz}{-2xz}.

Step 1 — First integral from dy/dzdy/dz

dy−2xy=dz−2xz  ⟹  dyy=dzz  ⟹  ln⁡y=ln⁡z+c1  ⟹  yz=C1.\dfrac{dy}{-2xy}=\dfrac{dz}{-2xz}\;\Longrightarrow\;\dfrac{dy}{y}=\dfrac{dz}{z}\;\Longrightarrow\;\ln y=\ln z+c_1\;\Longrightarrow\;\dfrac{y}{z}=C_1.

Step 2 — Second integral — use multipliers

Try multipliers (x,y,z)(x,y,z):

x dx+y dy+z dzx(y2+z2−x2)+y(−2xy)+z(−2xz)=x dx+y dy+z dzxy2+xz2−x3−2xy2−2xz2=x dx+y dy+z dz−x(x2+y2+z2).\dfrac{x\,dx+y\,dy+z\,dz}{x(y^2+z^2-x^2)+y(-2xy)+z(-2xz)}=\dfrac{x\,dx+y\,dy+z\,dz}{xy^2+xz^2-x^3-2xy^2-2xz^2}=\dfrac{x\,dx+y\,dy+z\,dz}{-x(x^2+y^2+z^2)}.

The numerator x dx+y dy+z dz=12d(x2+y2+z2)x\,dx+y\,dy+z\,dz=\tfrac{1}{2}d(x^2+y^2+z^2).

Combine with dz/(−2xz)dz/(-2xz):

12 d(x2+y2+z2)−x(x2+y2+z2)=dz−2xz.\dfrac{\tfrac{1}{2}\,d(x^2+y^2+z^2)}{-x(x^2+y^2+z^2)}=\dfrac{dz}{-2xz}.

Cross-multiply (cancel −x-x):

d(x2+y2+z2)2(x2+y2+z2)=dz2z.\dfrac{d(x^2+y^2+z^2)}{2(x^2+y^2+z^2)}=\dfrac{dz}{2z}.

Integrate:

12ln⁡(x2+y2+z2)=12ln⁡z+c  ⟹  x2+y2+z2z=C2.\dfrac{1}{2}\ln(x^2+y^2+z^2)=\dfrac{1}{2}\ln z+c\;\Longrightarrow\;\dfrac{x^2+y^2+z^2}{z}=C_2.

Step 3 — General solution

Answer

  Φ ⁣(yz,  x2+y2+z2z)=0     or equivalently     x2+y2+z2z=F ⁣(yz),  \boxed{\;\Phi\!\left(\dfrac{y}{z},\;\dfrac{x^2+y^2+z^2}{z}\right)=0\;\;\text{ or equivalently }\;\;\dfrac{x^2+y^2+z^2}{z}=F\!\left(\dfrac{y}{z}\right),\;}
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