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