- 2073 Bhadra · 5 marks
Euler's theorem verified for xⁿ tan⁻¹(y/x)
Short answer
Euler's theorem states that for a homogeneous function f of degree n, x f_x + y f_y = n f. For u = x^n arctan(y/x), it is homogeneous degree n, and direct calculation gives x u_x + y u_y = n u, verifying the theorem.
State Euler's Theorem for a homogeneous function of two independent variables and verify it for the function u = xⁿ tan⁻¹(y/x).
Euler's theorem
If f(x, y) is homogeneous of degree n (f(tx, ty) = tⁿ f(x, y)) and has continuous first partial derivatives, then
x ∂f/∂x + y ∂f/∂y = n f.
Verification for u = xⁿ tan⁻¹(y/x)
Degree: u(tx, ty) = tⁿ xⁿ tan⁻¹(ty/tx) = tⁿ u, so u is homogeneous of degree n. Euler's theorem predicts x u_x + y u_y = n u.
Partial derivatives (using d/dx tan⁻¹(y/x) = −y/(x² + y²) and d/dy tan⁻¹(y/x) = x/(x² + y²)):
∂u/∂x = n xⁿ⁻¹ tan⁻¹(y/x) + xⁿ ( −y/(x² + y²) )
∂u/∂y = xⁿ · x/(x² + y²) = xⁿ⁺¹/(x² + y²)
Then
x ∂u/∂x = n xⁿ tan⁻¹(y/x) − xⁿ⁺¹ y/(x² + y²)
y ∂u/∂y = xⁿ⁺¹ y/(x² + y²)
Adding, the fractions cancel:
x ∂u/∂x + y ∂u/∂y = n xⁿ tan⁻¹(y/x) = n u
This verifies Euler's theorem. ∎
✓ Answer checked by Claude Opus 5.5 (high effort)
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