Answer
$$
y^{\prime \prime}+4 y^{\prime}+4 y=0, \quad y(-1)=2, \quad y^{\prime}(-1)=1
$$
The general solution of the given initial value problem is
$$ y(t) =e^{-2t-2} ( 5 t +7) $$
$y(t) → 0$ as $t → ∞.$
Work Step by Step
$$
y^{\prime \prime}+4 y^{\prime}+4 y=0, \quad y(-1)=2, \quad y^{\prime}(-1)=1 \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+4r+4=(r+2)^{2}=0,
$$
so $r_{1}=r_{2}=-2$. Therefore one solution of Eq. (1) is $y_{1}=e^{-2t} $ .To find the general solution of Eq. (1), we need a second solution that is not a multiple of $y_{1}$. This second solution can be found, by using D’Alembert's method, by replacing $c$ by a function $v(t)$ and then trying to determine $ v(t)$ so that the product $v(t) y_{1}(t) $ is also a solution of Eq. (1).
Now we substitute $ y = v(t)y_{1}(t)$ in Eq. (1) and use the resulting equation to find $v(t)$.
$$ y=v(t) y_{1}(t)=v(t) e^{ -2t} \quad \quad (2)$$
we have
$$y^{\prime}=v^{\prime}(t) e^{-2t}-2 v(t) e^{-2t} \quad \quad (3)$$
and
$$
y^{\prime \prime}=v^{\prime \prime}(t) e^{ -2t}-4v^{\prime}(t) e^{-2t}+4v(t) e^{ -2 t} \quad \quad (4) $$
By substituting the expressions in Eqs. (2), (3), and (4) in Eq. (1) and collecting terms, we obtain
$$ v^{\prime \prime}(t)=0 $$
Therefore
$$v(t)=c_{1}t+c_{2} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants. Finally, substituting for $v(t)$ in Eq. (2), we obtain
$$ y(t)=c_{1}t e^{-2t} +c_{2} e^{-2t} \quad \quad (5)$$
The second term on the right side of Eq. (5) corresponds to the original solution $y_{1}(t) = e^{-2t}$, but the first term arises from a second solution, namely, $y_{2}(t) = t e^{-2 t}$. We can verify that these two solutions form a fundamental set by calculating their Wronskian: $W(y_{1}, y_{2})(t)\ne0$
Therefore $y_{1}(t) = e^{-2t}$, $y_{2}(t) = t e^{-2t}$ form a fundamental set of solutions of Eq. (1), and the general solution of that equation is given by
$$ y(t)=c_{1}t e^{-2t} +c_{2} e^{-2t} $$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
To apply the first initial condition, we set $t = -1 $ in Eq. (5); this gives
$$
y(0) = e^{2}(c_{2}-c_{1}) = 2. \quad \quad (6)
$$
For the second initial condition we must differentiate Eq. (5) as follows
$$
y^{\prime}(t) =e^{-2t}[(c_{1}-2c_{2})-2c_{1}t]
$$
and then set $t =-1 $. In this way we find that
$$ y^{\prime}(-1) =e^{2}(3c_{1}-2c_{2}) =1 \quad \quad (7)
$$
from eq.(6) and eq.(7), it follows that
$$
\left\{\begin{array}{ll}{e^{2}(c_{2}-c_{1}) } & {=2} \\ {e^{2}(3c_{1}-2c_{2}) } & {=1}\end{array}\right.
$$
This system of linear equations solves to $ c_{1} = \frac{5}{e^2} , c_{2} =\frac{7}{e^2} $.
Using these values of $ c_{1} = \frac{5}{e^2} , c_{2} =\frac{7}{e^2} $, in Eq. (5), we obtain
$$
\begin{split}
y(t) & =\frac{5}{e^2} t e^{-2t} +\frac{7}{e^2} e^{-2t}
\\
& = e^{-2t-2} ( 5 t +7)
\end{split}
$$
as the solution of the initial value problem (1).
Since $e^{-2t-2} $ goes to 0 as t → ∞, then $ y(t)$ goes to $0$ as t → ∞.
From the graph we can tell that the function $y(t) $ goes to $0$ as $t → ∞.$
