UPSC 2024 Maths Optional Paper 2 Q7a — 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 $(y-\phi)\phi_x+(\phi-x)\phi_y=x-y$ passing through the curve $\phi=0$ , $xy=1$ .

(Note: the problem also mentions a second condition — circle $x+y+\phi=0$ , $x^2+y^2+\phi^2=a^2$ — but these two conditions are mutually inconsistent for real $a$ ; the standard interpretation is to derive the surface from the first curve.)

Technique

Lagrange’s auxiliary equations; “add numerators” trick yields two independent first integrals; specialize using the given curve.

Solution

Step 1 — Lagrange auxiliary equations.

\frac{dx}{y-\phi}=\frac{dy}{\phi-x}=\frac{d\phi}{x-y}.

Step 2 — First integrals.

Add all numerators and denominators: denominator sum $(y-\phi)+(\phi-x)+(x-y)=0$ , so $dx+dy+d\phi=0$ :

x+y+\phi=c_1.

Multiply by $x,y,\phi$ respectively: denominator $x(y-\phi)+y(\phi-x)+\phi(x-y)=0$ , so $x\,dx+y\,dy+\phi\,d\phi=0$ :

x^2+y^2+\phi^2=c_2.

Step 3 — Apply the curve $\phi=0$ , $xy=1$ .

On this curve: $c_1=x+y$ and $c_2=x^2+y^2$ . Using $xy=1$ :

c_2=(x+y)^2-2xy=c_1^2-2,\quad\text{so }G(t)=t^2-2.

Integral surface: $x^2+y^2+\phi^2=(x+y+\phi)^2-2$ .

Expanding $(x+y+\phi)^2=x^2+y^2+\phi^2+2(xy+y\phi+x\phi)$ :

0=2(xy+y\phi+x\phi)-2\;\Rightarrow\;xy+y\phi+x\phi=1.

Answer

\boxed{\;xy+y\phi+x\phi=1.\;}