UPSC 2023 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Family of surfaces · PDEs · asked 7× in 13 yrs · Read the full method →

Question

By eliminating the arbitrary functions $f$ and $g$ from $z=f(x^2-y)+g(x^2+y)$ , form a partial differential equation.

Compute $z_x,z_y,z_{xx},z_{yy}$ in terms of $f',g',f'',g''$ ; the sum $f''+g''$ appears on both sides — eliminate via $z_{yy}$ .

Setup. Let $u=x^{2}-y$ and $v=x^{2}+y$ , so $z=f(u)+g(v)$ .

Step 1 — First-order partials.

z_x=f'(u)\cdot 2x+g'(v)\cdot 2x=2x\bigl(f'(u)+g'(v)\bigr),

z_y=f'(u)\cdot(-1)+g'(v)\cdot 1=g'(v)-f'(u).

Step 2 — Second-order partials.

$z_{xx}=\dfrac{d}{dx}[2x(f'+g')]=2(f'+g')+2x\cdot 2x(f''+g'')=2(f'+g')+4x^{2}(f''+g'')$ , i.e.

z_{xx}=\frac{z_x}{x}+4x^{2}(f''+g''). \tag{i}

z_{yy}=\dfrac{d}{dy}(g'-f')=g''+f''=f''+g''. \tag{ii}

Step 3 — Eliminate the arbitrary functions.

From (i): $4x^{2}(f''+g'')=z_{xx}-z_x/x$ . From (ii): $f''+g''=z_{yy}$ .

Substitute (ii) into (i):

z_{xx}-\frac{z_x}{x}=4x^{2}z_{yy}.

Equivalently, multiplying through by $x$ :

\boxed{\;x\,z_{xx}-z_x-4x^{3}z_{yy}=0.\;}