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Exercise 8.7 · Q4

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.

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Show u(x,y)=x2+y2x+yu(x,y)=\dfrac{x^2+y^2}{\sqrt{x+y}} is homogeneous of degree 32\tfrac32, then invoke Euler's Theorem directly for the required identity (with a direct-differentiation check).

Step 1. Test homogeneity.

u(λx,λy)=(λx)2+(λy)2λx+λy=λ2(x2+y2)λ x+y=λ2−1/2 x2+y2x+y=λ3/2u(x,y).u(\lambda x,\lambda y) = \frac{(\lambda x)^2+(\lambda y)^2}{\sqrt{\lambda x+\lambda y}} = \frac{\lambda^2(x^2+y^2)}{\sqrt\lambda\,\sqrt{x+y}} = \lambda^{2-1/2}\,\frac{x^2+y^2}{\sqrt{x+y}} = \lambda^{3/2}u(x,y).

So uu is homogeneous of degree 32\dfrac32.

Step 2. Apply Euler's Theorem directly. Since uu has continuous partial derivatives (away from x+y=0x+y=0) and is homogeneous of degree p=32p=\tfrac32, Euler's Theorem gives immediately

x∂u∂x+y∂u∂y=32 u(x,y).x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} = \frac32\,u(x,y).

Step 3. Direct-differentiation check (optional confirmation). u=(x2+y2)(x+y)−1/2u=(x^2+y^2)(x+y)^{-1/2}.

ux=2x(x+y)−1/2−x2+y22(x+y)3/2u_x=2x(x+y)^{-1/2}-\dfrac{x^2+y^2}{2(x+y)^{3/2}}, and by symmetry uy=2y(x+y)−1/2−x2+y22(x+y)3/2u_y=2y(x+y)^{-1/2}-\dfrac{x^2+y^2}{2(x+y)^{3/2}}. …

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