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=... .


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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