Question

In Problems $\mathbf{11}\mathbf{-}\mathbf{14}\mathbf{,}\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$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.

$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$

Short answer

The solution is $y=\frac{3}{2}{e}^x-\frac{1}{2}{e}^{-x}$.

Step-by-step solution

Step 1:

An initial value problem is a differential equation with some initial conditions.

Differentiate & 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 to x to get:

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

Differentiatewith respect to x to get:

$y^{\text{'}\text{'}}=c_1{e}^x+c_2{e}^{-x}$

Substitute initial value condition:

Substitute initial condition $y=c_1{e}^x+c_2{e}^{-x}$into get:

$$\begin{gathered} y\left(0\right)=1 \\ \Rightarrow 1=c_1{e}^0+c_2{e}^{-0} \end{gathered}$$

$\Rightarrow 1=c_1+c_2--\left(3\right)$

Now Substitute initial condition $y^{\text{'}}\left(0\right)=2$in equation (2) to get:

$$\begin{gathered} 2=c_1{e}^0-c_2{e}^{-0} \\ 2=c_1-c_2-----\left(4\right) \end{gathered}$$

Solve for constants:

Add equation (3) and equation (4) to get:

$2c_1=3$

$\Rightarrow c_1=\frac{3}{2}$

Substitute $c_1=\frac{3}{2}$in equation (3) implies:

$$\begin{gathered} C_2=1-\frac{3}{2} \\ =\frac{-1}{2} \end{gathered}$$

So, the solution of differential equation $y^{\text{'}\text{'}}-y=0$is $y=\frac{3}{2}{e}^x-\frac{1}{2}{e}^{-x}$.