Question

In Problems 29-36 solve the given initial-value problem.

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\left(0\right)\mathbf{=}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1} \end{gathered}$$

Short answer

$y=-\frac{1}{6}{e}^{-x}+\frac{1}{15}{e}^{2x}\frac{1}{10}{e}^{-3x}$

Step-by-step solution

Determine the initial value of differential equation 

The third order differential equation:

$y\text{'}\text{'}\text{'}+2y\text{'}\text{'}-5y\text{'}-6y=0$

Initial conditions:

$y\left(0\right)=0,y\text{'}\left(0\right)=0 \text{and} y\text{'}\text{'}\left(0\right)=1$

Consider$y=e^{mx}$as the solution of the differential equation,

Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$ and$y\text{'}\text{'}=m^2{e}^{mx}andy\text{'}\text{'}\text{'}=m^3{e}^{mx}$ into$y\text{'}\text{'}\text{'}+2y\text{'}\text{'}-5y\text{'}-6y=0$ to obtain the auxiliary equation.

$$\begin{gathered} m^{\text{3}}{e}^{\text{mx}}+2m^{\text{2}}{e}^{\text{mx}}-5m{e}^{\text{mx}}-6m{e}^{\text{mx}}=0 \\ {e}^{\text{mx}}\left(m^{\text{3}}+2m^{\text{2}}-5m-6\right)=0 \end{gathered}$$

For any real$x,e^{mx}\ne 0$, we have

$$\begin{gathered} m^{\text{3}}+2m^2-5m-6=0 \\ \left(m+1\right)\left(m^{\text{2}}+m+6\right)=0 \\ \left(m+1\right)\left(m-2\right)\left(m+3\right)=0 \end{gathered}$$

Roots are:

$m_1=-1,m_2=2 \text{and} m_3=-3$

Find the general solution

$y=k_1{e}^{\text{- x}}+k_2{e}^{\text{2x}}+k_3{e}^{\text{- 3x}}······\left(1\right)$

Arbitrary constant:$k_1,k_{2,}{k}_3$

Now we have to apply the given conditions in the general solution of our given differential equations as the following technique.

We have to apply with the point in the general solution as

role="math" localid="1663904943758" $$\begin{gathered} 0=k_1{e}^0+k_2{e}^0+k_3{e}^{\text{0}} \\ {k}_1+k_2+k_3=0 \end{gathered}$$

After that, we have to take the first derivative for the general solution of the given differential equation as

$$\begin{gathered} y\text{'}=-k_1{e}^{\text{- x}}+2k_2{e}^{\text{2x}}-3k_3{e}^{\text{- 3x}} \\ 0=-k_1{e}^0+2k_2{e}^0-3k_3{e}^0 \\ -k_1+2k_2-3k_3=0 \end{gathered}$$

After that, we have to take the second derivative for the general solution (1) of the given differential equation as.

$y\text{'}\text{'}=k_1{e}^{-x}+4k_2{e}^{-6x}+9k_3{e}^{-3x}$

Apply with the point:

$$\begin{gathered} \left(x,y\text{'}\text{'}\right)=\left(0,1\right) \\ 1=k_1{e}^0+4k_2{e}^0+9k_3{e}^0 \\ {k}_1+4k_2+9k_3=1 \end{gathered}$$

Then by solving the three equations, we find that

$$\begin{gathered} k_1=-\frac{1}{6},k_2=\frac{1}{15} \text{and} k_3=\frac{1}{10} \\ y=-\frac{1}{6}{e}^{-x}+\frac{1}{15}{e}^{2x}\frac{1}{10}{e}^{-3x} \end{gathered}$$

The solution of the differential equation$y\text{'}\text{'}\text{'}+2y\text{'}\text{'}-5y\text{'}-6y=0$

$y=-\frac{1}{6}{e}^{-x}+\frac{1}{15}{e}^{2x}\frac{1}{10}{e}^{-3x}$

Hence the general solution is$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{6}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{15}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\frac{\mathbf{1}}{\mathbf{10}}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}$