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Question 90 of 99

Q.If u(x,y)=x2+y2x+yu(x, y)=\dfrac{x^2+y^2}{\sqrt{x+y}}, prove that x∂u∂x+y∂u∂y=32ux\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=\dfrac32 u.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2024Subjective· 3mImportance★★★★★
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Recognises uu as a homogeneous function of degree 3/23/2 and applies Euler's theorem directly, then confirms by direct differentiation.

  1. Degree check. u(x,y)=x2+y2x+yu(x,y)=\dfrac{x^2+y^2}{\sqrt{x+y}}: scaling x→tx, y→tyx\to tx,\,y\to ty gives u(tx,ty)=t2(x2+y2)tx+y=t2−1/2u(x,y)=t3/2u(x,y)u(tx,ty)=\dfrac{t^2(x^2+y^2)}{\sqrt t\sqrt{x+y}}=t^{2-1/2}u(x,y)=t^{3/2}u(x,y) — so uu is homogeneous of degree n=32n=\dfrac32.
  2. Euler's theorem states that for a homogeneous function of degree nn: x∂u∂x+y∂u∂y=nux\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=nu. Here n=32n=\dfrac32, giving the required result directly.
  3. Direct check. u=(x2+y2)(x+y)−1/2u=(x^2+y^2)(x+y)^{-1/2}. ux=2x(x+y)−1/2−12(x2+y2)(x+y)−3/2u_x=2x(x+y)^{-1/2}-\dfrac12(x^2+y^2)(x+y)^{-3/2}; by symmetry uy=2y(x+y)−1/2−12(x2+y2)(x+y)−3/2u_y=2y(x+y)^{-1/2}-\dfrac12(x^2+y^2)(x+y)^{-3/2}. …

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