UPSC 2013 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

Form a partial differential equation by eliminating the arbitrary functions $f$ and $g$ from $z=yf(x)+xg(y)$ .

Form $xz_x$ and $yz_y$ to isolate $f',g'$ pieces from $f,g$ pieces; combine to use $z=yf+xg$ as an algebraic identity.

Strategy. Compute the relevant partials, then combine to eliminate $f, f', g, g'$ .

From $z=yf(x)+xg(y)$ :

z_x=yf'(x)+g(y),\qquad z_y=f(x)+xg'(y),\qquad z_{xy}=f'(x)+g'(y).

Multiply $z_x$ by $x$ :

xz_x=xyf'(x)+xg(y).

Subtract $z=yf+xg$ :

xz_x-z=xyf'(x)+xg(y)-yf(x)-xg(y)=xyf'(x)-yf(x). \tag{i}

Multiply $z_y$ by $y$ :

yz_y=yf(x)+xyg'(y).

Subtract $z$ :

yz_y-z=yf(x)+xyg'(y)-yf(x)-xg(y)=xyg'(y)-xg(y). \tag{ii}

Add (i) and (ii):

xz_x+yz_y-2z=xy\bigl(f'(x)+g'(y)\bigr)-\bigl(yf(x)+xg(y)\bigr)=xy\,z_{xy}-z.

(Using $f'+g'=z_{xy}$ and $yf+xg=z$ .)

xz_x+yz_y-2z=xyz_{xy}-z\;\Longrightarrow\;xz_x+yz_y-z=xyz_{xy}.

\boxed{\;xy\,\frac{\partial^{2}z}{\partial x\partial y}=x\,\frac{\partial z}{\partial x}+y\,\frac{\partial z}{\partial y}-z.\;}