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Exercise 8.7 · Q5

Q.If v(x,y)=log⁡(x2+y2x+y)v(x,y)=\log\left(\dfrac{x^2+y^2}{x+y}\right), prove that x∂v∂x+y∂v∂y=1x\dfrac{\partial v}{\partial x}+y\dfrac{\partial v}{\partial y}=1.

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v=log⁡fv=\log f where f=x2+y2x+y=evf=\dfrac{x^2+y^2}{x+y}=e^v is itself homogeneous of degree 11; apply Euler's Theorem to ff, then convert the identity from ff to vv via the chain rule (exactly the trick of Example 8.22).

Step 1. Identify the homogeneous "core". Let f(x,y)=x2+y2x+y=evf(x,y)=\dfrac{x^2+y^2}{x+y}=e^v (since v=log⁡fv=\log f). Test homogeneity of ff:

f(λx,λy)=λ2(x2+y2)λ(x+y)=λ f(x,y).f(\lambda x,\lambda y) = \frac{\lambda^2(x^2+y^2)}{\lambda(x+y)} = \lambda\,f(x,y).

So ff is homogeneous of degree 11.

Step 2. Apply Euler's Theorem to ff. xfx+yfy=1⋅f=fx f_x+yf_y = 1\cdot f = f.

Step 3. Relate fx,fyf_x,f_y to vx,vyv_x,v_y via the chain rule. Since f=evf=e^v: fx=evvxf_x=e^v v_x and fy=evvyf_y=e^v v_y (chain rule on e(⋅)e^{(\cdot)}).

Step 4. Substitute into the Euler identity from Step 2. …

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