Name: Higher Order Derivative
Uploaded: 2019-07-19
Duration: 4 min 20 s
Description: Did you know that functions can be used as exponents? We find derivatives of that function by logarithm. What is a higher-order derivative? Watch this video...

Question

If

y = A e^{- k t} \cos (p t + c)

Prove that

\frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0

where

n^{2} = p^{2} + k^{2}

Accepted Answer

We have, $y = A e^{- k t} \cos (p t + c) \dots (1)$

Differentiating $y$ with respect to $t,$ we get

$\frac{d y}{d t} = - k A e^{- k t} \cos (p t + c) - p A e^{- k t} \sin (p t + c)$

$= - k y - p A e^{- k t} \sin (p t + c)$ [From $(1)$ ]

$y = A e^{- k t} \cos (p t + c)$

Differentiating $\frac{d y}{d t}$ with respect to $n^{2} = p^{2} + k^{2}$ , we get

$\frac{d^{2} y}{d t^{2}} = - k \frac{d y}{d t} + p k A e^{- k t} \sin (p t + c) - p^{2} A e^{- k t} \cos (p t + c)$

$= - k \frac{d y}{d t} + k (- k y - \frac{d y}{d t}) - p^{2} y$ [From $(1)$ and $(2)$ ]

$= - k \frac{d y}{d t} - k^{2} y - k \frac{d y}{d t} - p^{2} y$

$= - 2 k \frac{d y}{d t} - (k^{2} + p^{2}) y$

$\Rightarrow \frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + (k^{2} + p^{2}) y = 0$

$\Rightarrow \frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0$ , where $n^{2} = p^{2} + k^{2}$

Hence $\frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0$ , where $n^{2} = p^{2} + k^{2}$ .

If $y = A e^{- k t} \cos (p t + c)$ Prove that $\frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0$ where $n^{2} = p^{2} + k^{2}$

Important Questions on Higher Order Derivatives

Learn and Prepare for any exam you want !

If y=Ae-ktcos(pt+c) Prove that d2ydt2+2kdydt+n2y=0 where n2=p2+k2

Important Questions on Higher Order Derivatives

Learn and Prepare for any exam you want !

If $y = A e^{- k t} \cos (p t + c)$ Prove that $\frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0$ where $n^{2} = p^{2} + k^{2}$