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Question
If v(x, y) = `log((x^2 + y^2)/(x + y))`, prove that `x (del"v")/(delx) + y (del"u")/(dely) = 1`
Solution
v(x, y) = `log((x^2 + y^2)/(x + y))`
Change into exponential function
Let `"e"^"v" = (x^2 + y^2)/(x + y) = "f"(x, y)`
f(x, y) = `(lambda^2x^2 + lambda^2y^2)/(lambdax + lambday)`
= `(lambda^2(x^2 + y^2))/(lambda(x + y))`
By Euler's Theorem
`x (del"f")/(delx) + y (del"f")/(dely) = 1 xx "f" = "f"`
`x (del"f")/(delx) "e"^"v" + y del/(dely)` = ev exists.
`x "e"^"v" (del"f")/(delx) + y "e"^"v" (del"f")/(dely)` = ev
`x (del"f")/(delx) + y (del"f")/(dely) = "e"^"v"/"e"^"v"` = 1
Hence proved.
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