Question

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

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{12}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{36}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0} \\ \mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{7} \end{gathered}$$

Short answer

$y=\frac{5}{36}-\frac{5}{36}{e}^{\text{- 6x}}+\frac{1}{6}x{e}^{\text{- 6x}}$

Step-by-step solution

Step 1: Determine the initial value of differential equation

The third order differential equation:

$y\text{'}\text{'}\text{'}+12y\text{'}\text{'}+36y\text{'}=0$

Initial conditions:

$y\left(0\right)=0,y\text{'}\left(0\right)=1 \text{and} y\text{'}\text{'}\left(0\right)=-7$

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 $y\text{'}\text{'}\text{'}+12y\text{'}\text{'}+36y\text{'}=0$to obtain the auxiliary equation.

$$\begin{gathered} m^3{e}^{mx}+12m^2{e}^{mx}+36e^{mx}=0 \\ {e}^{mx}\left(m^3+12m^2+36m\right)=0 \end{gathered}$$

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

$$\begin{gathered} m^3+12m^2+36m=0 \\ m\left(m^3+12m^2+36m\right)=0 \\ m\left(m+6\right)\left(m+6\right)=0 \end{gathered}$$

Roots are:

$m_1=0 \text{and} m_{2,3}=-6$

Real and repeated:

$m_{2,3}=-6$

Find the additional linearly independent solution

Then we need to have an additional linearly independent solution for this repeated root, can obtain the general solution of our differential equation

$$\begin{gathered} y=c_1{e}^{\text{0}}+n{e}^{\text{- 6x}}\left(k+sx\right) \\ =c_1+c_2{e}^{\text{- 6x}}+c_{3}x{e}^{\text{- 6x}}······\left(1\right) \end{gathered}$$

Arbitrary constants:

$c_2=n\times k,c_3=n\times s$

Fourth, 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

$$\begin{gathered} 0=c_1+c_2{e}^{\text{0}}+c_3\left(0\right) \\ {c}_1+c_2=0 \end{gathered}$$

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

$y\text{'}=-6c_2{e}^{\text{- 6x}}+c_3{e}^{\text{- 6}x}-6c_{3}x{e}^{\text{- 6}x}$

Apply with the point $\left(x,y\text{'}\right)=\left(0,1\right)$

$$\begin{gathered} 1=-6c_2{e}^{\text{0}}+c_3{e}^0-6c_3\left(0\right) \\ -6c_2+c_3=1 \end{gathered}$$

After that, we have to take the second derivative for the general solution (4) of the given differential equation as

$y\text{'}\text{'}=36c_2{e}^{\text{- 6x}}-12c_3{e}^{-6x}+36c_{3}x{e}^{-6x}$

Apply the point$\left(x,y\text{'}\text{'}\right)=\left(0,-7\right)$

$36c_2-12c_3=-7$

Then by solving the three equations $c_1,c_2\text{and}{c}_3$we find that

$c_1=\frac{5}{36},c_2=-\frac{5}{36} \text{and} c_3=\frac{1}{6}$

General solution of our given differential equation,

$y=\frac{5}{36}-\frac{5}{36}{e}^{\text{- 6x}}+\frac{1}{6}x{e}^{\text{- 6x}}$

Hence, the general solution is$\mathit{y}\mathbf{=}\frac{\mathbf{5}}{\mathbf{36}}\mathbf{-}\frac{\mathbf{5}}{\mathbf{36}}{\mathit{e}}^{\mathbf{\text{- 6x}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{6}}\mathit{x}{\mathit{e}}^{\mathbf{\text{- 6x}}}$