If y-ae2x-be-x- then show that d2ydx2-dydx-2y-0-

y=ae^2x+be^ x dydx=2ae^2x be^ x d^2ydx^2=4ae^2x+be^ x Putting these values in LHS, we get, d^2ydx^2 dydx 2y =(4ae^2x+be^ x) (2ae^2x be^ ...

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Higher Order Derivatives

If y=ae2x+b...

Question

If $y = a e^{2 x} + b e^{- x}$ , then show that $\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 y = 0$ .

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y = a e^{2 x} + b e^{- x}

\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 y = 0

\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 y = 0

y = x^{2} e^{x}

\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 (x + 1) e^{x} = 0

y = e^{- x} cos 2 x

y = f (x)

x

\frac{d^{2} x}{d y^{2}} = - {(\frac{d y}{d x})}^{- 3} . \frac{d^{2} y}{d x^{2}}

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