Question

In Problems 11-14, $\mathbf{11}\mathbf{-}\mathbf{14}\mathbf{,}\mathbf{}\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$is a two-parameter family of solutions of the second-order DE ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

$\mathit{y}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{5}$

Short answer

The required solution is $y=5e^{-x-1}$

Step-by-step solution

Differentiate and Evaluate

Given a differential equation $y=c_1{e}^x+c_2{e}^{-x}\dots \dots \left(1\right)$

Now differentiate equation (1) with respect toxto get:

$y^{\text{'}}=c_1{e}^x-c_2{e}^{-x}\dots \dots \text{(2)}$

Differentiatewith respect toxto get:

$y^{\text{'}\text{'}}=c_1{e}^x+c_2{e}^{-x}······\left(3\right)$

Substitute initial value condition

Substitute initial condition $y(-1)=5$in $y=c_1{e}^x+c_2{e}^{-x}$into get:

$$\begin{gathered} y(-1)=5\backslash \\ \Rightarrow 5=c_1{e}^{-1}+c_2{e}^1 \end{gathered}$$

$\Rightarrow 5=\frac{c_1}{e}+e{c}_2······\left(4\right)$

Now Substitute initial condition$y^{\text{'}}(-1)=-5$ in equation (2) to get:

$$\begin{gathered} -5=c_1{e}^{-1}-c_2{e}^1 \\ -5=\frac{c_1}{e}-e{c}_2······\left(5\right) \end{gathered}$$

Solve for constants

Add equation (4) and equation (5) to get:

$\frac{2c_1}{e}=5-5$

$$\begin{gathered} \Rightarrow \frac{2c_1}{e}=0 \\ \Rightarrow c_1=0 \end{gathered}$$

Substitute $c_1=0$in equation (3) implies:

$$\begin{gathered} c_2=5 \\ {c}_2=\frac{5}{e} \end{gathered}$$

Substitute $c_1$and $c_2$values in equation (1),

$$\begin{gathered} y=\frac{5}{e}{e}^{-x} \\ y=5e^{-x-1} \end{gathered}$$.

Hence, the required solution is$\mathit{y}\mathbf{=}\mathbf{5}\mathbf{}{\mathit{e}}^{\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{1}}$.