Given that y=c1e3t+c2e−3t a solution to the differential equation y′′−9y=0, where c1 and c2 are arbitrary constants, find a function y that satisfies the conditions: y′′−9y=0, y(0)=7, limt→+[infinity]y(t)=0. Give your answer as y=... .

Answer:

$y = 7e^{-3t}$

Step-by-step explanation:

$y = C_1e^{3t} + C_2e^{-3t}$

$y(0) = 7$ implies that when $t=0$ , $y=7$ .

$7 = C_1e^{3\times0) + C_2e^{-3\times0}$

$7 = C_1+C_2$

For the condition $\lim\limits_{t \to +\infty} y(t) = 0$ , as $t\to+\infty$ , $y\to0$ .

In the expression for $y$ , as $t\to+\infty$ , the term containing $C_2$ vanishes. Hence,

$0 = C_1e^{3t}$

$C_1 = 0$

Since $7 = C_1+C_2$

$7 = 0+C_2$

$C_2 = 7$

Substituting for $C_1$ and $C_2$ in $y$ ,

$y = 7e^{-3t}$

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