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a\overset{\rightarrow}{a} ഉം b\overset{\rightarrow}{b} ഉം കോൺസ്റ്റന്റ് വെക്ടറുകളും r=ae5t+be5t\overset {\rightarrow}{r}=\overset{\rightarrow}{a}e^{5t}+ \overset{\rightarrow}{b}e^{-5t} ഉം ആണെങ്കിൽd2rdt225r \frac{d^2r}{dt^2}-25r ആണ്

A4

B3

Czero

D5

Answer:

C. zero

Read Explanation:

Given:
r=ae5t+be5t\vec{r} = \vec{a}e^{5t} + \vec{b}e^{-5t}

First derivative

drdt=5ae5t5be5t\frac{d\vec{r}}{dt} = 5\vec{a}e^{5t} - 5\vec{b}e^{-5t}

Second derivative

d2rdt2=25ae5t+25be5t\frac{d^2\vec{r}}{dt^2} = 25\vec{a}e^{5t} + 25\vec{b}e^{-5t}

Compute (d2rdt225r)( \frac{d^2\vec{r}}{dt^2} - 25\vec{r} )

=(25ae5t+25be5t)25(ae5t+be5t)= (25\vec{a}e^{5t} + 25\vec{b}e^{-5t}) - 25(\vec{a}e^{5t} + \vec{b}e^{-5t})
=25ae5t+25be5t25ae5t25be5t= 25\vec{a}e^{5t} + 25\vec{b}e^{-5t} - 25\vec{a}e^{5t} - 25\vec{b}e^{-5t}
= 0


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