Step 1 : Concept:
This question tests Euler's Theorem for homogeneous functions and its extensions for implicit homogeneous functions. Step 2 : Key Formulas and Approach:
1. Euler's Theorem: If $u(x,y)$ is homogeneous of degree $n$, then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$.
2. Extension: If $f(u)$ is a homogeneous function of degree $n$, then:
\[
x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{n f(u)}{f'(u)}
\]
Step 3 : Step-by-step Explanation:
• Item A: $u = e^{x^2 + y^2} \implies \log u = x^2 + y^2$.
Let $f(u) = \log u$, which is homogeneous of degree $n = 2$.
$f'(u) = \frac{1}{u}$.
\[
x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2 \log u}{1/u} = 2u \log u
\]
Matches with IV.
• Item B: $u(x,y) = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right)$.
Here $u(tx, ty) = \sin^{-1}\left(\frac{tx}{ty}\right) + \tan^{-1}\left(\frac{ty}{tx}\right) = t^0 u(x,y)$.
Degree of homogeneity $n = 0$. By Euler's theorem:
\[
x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \cdot u = 0
\]
Matches with I.
• Item C: $u = \cos^{-1}\left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right) \implies \cos u = \frac{x+y}{\sqrt{x}+\sqrt{y}}$.
$f(u) = \cos u$ is homogeneous of degree $n = 1 - 1/2 = 1/2$.
$f'(u) = -\sin u$.
\[
x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\frac{1}{2}\cos u}{-\sin u} = -\frac{1}{2} \cot u
\]
Matches with II.
• Item D: $u = x^2 + y^2$ is homogeneous of degree $n = 2$.
By Euler's theorem:
\[
x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2u
\]
Matches with III. Step 4 : Final Answer:
The correct matching is A-IV, B-I, C-II, D-III, which corresponds to option (D).
