Question:medium

If $u(x, y) = \sin^{-1} \frac{x}{y} + \tan^{-1} \frac{y}{x}$ then $xu_x + yu_y =$

Show Hint

Whenever you see a function where the variables $x$ and $y$ only appear as ratios (like $x/y$ or $y/x$), the function is always homogeneous of degree zero. For such functions, $xu_x + yu_y$ will always be zero!
Updated On: Jul 1, 2026
  • $u'(x, y)$
  • $0$
  • $1$
  • $u(x, y)$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Check for Homogeneity: A function $f(x, y)$ is homogeneous of degree $n$ if $f(tx, ty) = t^n f(x, y)$. Let's test our function $u(x, y)$: $$u(tx, ty) = \sin^{-1} \left( \frac{tx}{ty} \right) + \tan^{-1} \left( \frac{ty}{tx} \right)$$ $$u(tx, ty) = \sin^{-1} \left( \frac{x}{y} \right) + \tan^{-1} \left( \frac{y}{x} \right)$$ $$u(tx, ty) = t^0 \cdot u(x, y)$$

Step 2: Identify the Degree ($n$): Since $u(tx, ty) = t^0 u(x, y)$, the function is a homogeneous function of degree $n = 0$.

Step 3: Apply Euler's Theorem: Euler's Theorem states that for a homogeneous function $u$ of degree $n$: $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = nu$$ Substituting our degree $n = 0$: $$xu_x + yu_y = 0 \cdot u = 0$$ Thus, the value of the expression is always 0.
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