## Elementary Differential Equations and Boundary Value Problems 9th Edition

$$y^{\prime \prime}+4 y^{\prime}+4 y=0, \quad y(-1)=2, \quad y^{\prime}(-1)=1$$ The general solution of the given initial value problem is $$y(t) =e^{-2t-2} ( 5 t +7)$$ $y(t) → 0$ as $t → ∞.$
$$y^{\prime \prime}+4 y^{\prime}+4 y=0, \quad y(-1)=2, \quad y^{\prime}(-1)=1 \quad (1)$$ 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 $t → ∞.$