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Question
Prove that g(x, y) = `x log(y/x)` is homogeneous What is the degree? Verify Eulers Theorem for g
Solution
g(x, y) = `x log(y/x)`
g(tx, ty) = `"t"x log(("t"y)/("t"x))`
g is a homogeneous function of degree 1.
∴ By Euler’s Theorem,
`x (del"g")/(delx) + y (del"g")/(dely)` = g
Verification:
g(x, y) = `xlog(y/x)`
= `x (logy - log x)`
= `x log y - x log x`
`(del"g")/(delx) = logy - logx - x xx 1/x`
= `log y - log x - 1`
`x (del"g")/(delx) = x log y - x log x - x`
`(del"")/(dely) = x xx 1/y`
`y (delg")/(dely)` = x
`x (del"g")/(delx) + y (del"g")/(dely) = x log y - x log x - x + x`
= `x log (y/x)`
= g
`x (del"g")/(delx) + y (del"g")/(dely)` = g
Hence verified.
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