Question

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 1

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 $y$,

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

Answer 2

The answer is $y = 7e^{-3t}$

Since $y = c_{1}e^{3t} + c_{2}e^{-3t}$ is a solution to the differential equation y" - 9y = 0, and given that y(0) = 7, $\lim_{t \to +\infty} y(t) = 0$.

The equation

So,  $y = c_{1}e^{3t} + c_{2}e^{-3t}$

$y(0) = c_{1}e^{3(0)} + c_{2}e^{-3(0)}\\7 = c_{1}e^{0} + c_{2}e^{0}\\7 = c_{1} + c_{2}$

The constants

Also,  $\lim_{t \to +\infty} y(t) = 0$

So, $y(\infty) = c_{1}e^{3(+\infty)} + c_{2}e^{-3(+\infty)}\\0 = c_{1}e^{\infty} + c_{2}e^{-\infty}\\0 = c_{1}\infty + c_{2}(0)\\0 = c_{1}\infty + 0\\0/\infty = c_{1}\\c_{1} = 0$

Since c₁ = 0, substituting it into c₁ + c₂ = 7, we have

c₁ + c₂ = 7

0 + c₂ = 7

c₂ = 7

The value of y

So, substituting c₁ = 0 and c₂ = 7 into y, we have

$y = c_{1}e^{3t} + c_{2}e^{-3t}\\y = (0)e^{3t} + 7e^{-3t}\\y = 0 + 7e^{-3t}\\y = 7e^{-3t}$

So, the answer is $y = 7e^{-3t}$

Learn more about differential equations here:

https://brainly.com/question/18760518