UPSC Maths 2025 Paper 1 Q3c-ii — Solution

10 marks · Section A

Question

If $u(x, y) = x\,f\!\left(\dfrac{y}{x}\right) + g\!\left(\dfrac{y}{x}\right)$ , where $f$ and $g$ are arbitrary functions, then show that

I. $x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = x\,f\!\left(\dfrac{y}{x}\right)$ ,
II. $x^2\dfrac{\partial^2 u}{\partial x^2} + 2xy\dfrac{\partial^2 u}{\partial x\,\partial y} + y^2\dfrac{\partial^2 u}{\partial y^2} = 0$ .

Technique

Use Euler’s theorem on homogeneous functions: $x f(y/x)$ is homogeneous of degree $1$ and $g(y/x)$ is homogeneous of degree $0$ . Apply Euler’s theorem to each part (direct differentiation also confirms Part I).

Solution

Let $t = y/x$ , so $\dfrac{\partial t}{\partial x} = -\dfrac{y}{x^2}$ , $\dfrac{\partial t}{\partial y} = \dfrac{1}{x}$ .

Part I.

First derivatives: $u_x = f(t) + x f'(t)\cdot\left(-\frac{y}{x^2}\right) + g'(t)\cdot\left(-\frac{y}{x^2}\right) = f(t) - \frac{y}{x}f'(t) - \frac{y}{x^2}g'(t).$ $u_y = x f'(t)\cdot\frac{1}{x} + g'(t)\cdot\frac{1}{x} = f'(t) + \frac{1}{x}g'(t).$

Then $x u_x + y u_y = x f(t) - y f'(t) - \frac{y}{x}g'(t) + y f'(t) + \frac{y}{x}g'(t) = x f(t).$ $x u_x + y u_y = x\,f\!\left(\frac yx\right).\qquad\text{(I proved)}$

Euler view: write $u = P + Q$ with $P=x f(y/x)$ (degree 1) and $Q=g(y/x)$ (degree 0). Euler’s theorem gives $xP_x+yP_y=1\cdot P$ and $xQ_x+yQ_y=0\cdot Q=0$ . Adding: $xu_x+yu_y=P=xf(y/x)$ . ✓

Part II.

Apply Euler’s theorem at second order. For a function homogeneous of degree $n$ , $x^2\,\phi_{xx}+2xy\,\phi_{xy}+y^2\,\phi_{yy}=n(n-1)\phi.$

For $P=xf(y/x)$ , $n=1$ : $\;x^2P_{xx}+2xyP_{xy}+y^2P_{yy}=1\cdot0\cdot P=0.$
For $Q=g(y/x)$ , $n=0$ : $\;x^2Q_{xx}+2xyQ_{xy}+y^2Q_{yy}=0\cdot(-1)\cdot Q=0.$

Adding (since $u=P+Q$ and the operator is linear): $x^2u_{xx}+2xy\,u_{xy}+y^2u_{yy}=0.\qquad\text{(II proved)}$

Answer

$\boxed{\;x u_x + y u_y = x\,f\!\left(\frac yx\right),\qquad x^2u_{xx}+2xy\,u_{xy}+y^2u_{yy}=0.\;}$