Math, asked by dhhdhdhdh, 1 year ago

ify=xlog(x/a+bx)then prove that x3d2y/dx2=(xdy/dx-y)2✏✏✏

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Answered by Anonymous
6
y = x Log [x/(a+bx)]    ---- (1)
Let   x/(a+bx) = z      ---- (2)

To prove  x³ d²y/dx² = (x y' - y)²

y' = Log z + x * (a+bx)/x * a/(a+bx)²
   = Log z + a/(a+bx)   --- (3)

y'' = 1/z * a/(a+bx)² - ab/(a+bx)²

LHS = x³ y'' = a x² / (a+bx) - ab x³/(a+bx)²
         = ax² / (a+bx)² * [a + bx - b x] = a²x² / (a+bx)²

RHS = (x y' - y)²
        = [x Log z + ax/(a+x) - x Log z ]²
        = a²x²/(a+x)²

LHS = RHS Proved.
Answered by silu12
4
hii
here is ur ans....

y=x log (x/a+bx)

let x/a+bx=t

y=xlog t

dy/dx =logt+x.1/t

d^2y/dx^2=1/t+1/t+x×(-1/t^2)

putting the value of t ....then u found ur prove
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