9E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems $\mathbf{1}\mathbf{-}\mathbf{14}$ , solve the given initial value problem using the method of Laplace transforms. 

$\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{z}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{z}\mathbf{=}\mathbf{21}{\mathbf{e}}^{\mathbf{t}\mathbf{-}\mathbf{1}}\mathbf{,} \mathbf{z}\left(\mathbf{1}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,} \mathbf{z}\mathbf{\text{'}}\left(\mathbf{1}\right)\mathbf{=}\mathbf{9}$

Step-by-Step Solution

Answer

The Initial value for $z\text{'}\text{'}+5z\text{'}-6z=21e^{t-1}$ is $z\left(t\right)=3\left(t-1\right)e^{t-1}-e^{-6\left(t-1\right)}$

Step 1: Define the Laplace Transform

• The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain. 
• In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Step 2: Determine the initial value of Laplace transform

Shift the initial conditions to $t=0$ by defining a new function:

$$\begin{gathered} y\left(t\right)=z\left(t+1\right) \\ y\text{'}\left(t\right)=z\text{'}\left(t+1\right) \\ y\text{'}\text{'}\left(t\right)=z\text{'}\text{'}\left(t+1\right) \end{gathered}$$

Replace $t$ by $t+1$  in the condition

$$\begin{gathered} z\text{'}\text{'}\left(t\right)+5z\text{'}\left(t\right)-6z\left(t\right)=21e^{t-1} \\ \Rightarrow z\text{'}\text{'}\left(t+1\right)+5z\text{'}\left(t+1\right)-6z\left(t+1\right)=21e^{\left(t+1\right)-1} \end{gathered}$$

$\Rightarrow z\text{'}\text{'}\left(t+1\right)+5z\text{'}\left(t+1\right)-6z\left(t+1\right)=21e^t$

Substitute $y\left(t\right)=z\left(t+1\right)$

Solve for initial condition.

$$\begin{gathered} y\left(0\right)=z\left(0+1\right) \\ =z\left(1\right) \\ =-1 \end{gathered}$$

Solve for differentiation of initial condition.

$$\begin{gathered} y\text{'}\left(0\right)=z\text{'}\left(0+1\right) \\ =z\text{'}\left(1\right) \\ =9 \end{gathered}$$

Simplify the equation as:

$$\begin{gathered} y\text{'}\text{'}\left(t\right)+5y\text{'}\left(t\right)-6y\left(t\right)=21e^t \\ y\left(0\right)=-1 \\ y\text{'}\left(0\right)=9 \end{gathered}$$

Define $\mathcal{L}\left\{y\right\}\left(s\right)=Y\left(s\right)$

sing the properties listed below, take the Laplace transform of the equation

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\right\}\left(s\right)=s\mathcal{L}\left\{y\right\}\left(s\right)-y\left(0\right) \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}\left(s\right)=s^2\mathcal{L}\left\{y\right\}\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right) \end{gathered}$$

$$\begin{gathered} \mathcal{L}\left\{e^{at}\right\}\left(s\right)=\frac{1}{s-a} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+5L\left\{y\text{'}\right\}-6L\left\{y\right\}=21L\left\{e^t\right\} \end{gathered}$$

Substitute the properties into the equation.

$\left[s^{2}Y-sy\left(0\right)-y\text{'}\left(0\right)\right]+5\left[sY-y\left(0\right)\right]-6Y=\frac{21}{s-1}$

Substitute the initial conditions:

$y\left(0\right)=-1 and y\text{'}\left(0\right)=9$

$\left(s^{2}Y+s-9\right)+5\left(sY+1\right)-6Y=\frac{21}{s-1}$

Distribute and simplify:

$\Rightarrow s^{2}Y+5sY-6Y+s-4=\frac{21}{s-1}$

Isolate the Y variable.

$$\begin{gathered} Y\left(s^2+5s-6\right)=\frac{21}{s-1}-s+4 \\ \Rightarrow Y\left(s^2+5s-6\right)=\frac{21+\left(-s+4\right)\left(s-1\right)}{s-1} \\ \Rightarrow Y=\frac{-s^2+5s+17}{\left(s-1\right)\left(s^2+5s-6\right)} \end{gathered}$$

Find the partial fraction expansion

Because $s-1$ is a repeated factor of ${\left(s-1\right)}^2\left(s+6\right)$ ,we include $\left(s-1\right)$ and ${\left(s-1\right)}^2$

$Y=\frac{-s^2+5s+17}{\left(s-1\right)\left(s-1\right)\left(s+6\right)}$

$\Rightarrow \frac{-s^2+5s+17}{{\left(s-1\right)}^2\left(s+6\right)}=\frac{A}{s-1}+\frac{B}{{\left(s-1\right)}^2}+\frac{C}{s+6}$

Combine the fractions to equate the numerators.

$-s^2+5s+17=A\left(s-1\right)\left(s+6\right)+B\left(s+6\right)+C{\left(s-1\right)}^2$

Solve for variables by setting values of S

$$\begin{gathered} s=1\Rightarrow 21=A·0+7B+C·0=7B\Rightarrow B=3 \\ s=-6\Rightarrow -49=A·0+3·0+49C\Rightarrow C=-1 \end{gathered}$$

$s=0\Rightarrow 17=-6A+3·6-1·1\Rightarrow A=0$

Substitute the values  A,B,C of into partial fraction expansion

Using the properties listed below take the inverse Laplace transform  to obtain the solution $y\left(t\right)$

$$\begin{gathered} c^{-1}\left\{\frac{1}{s-a}\right\}=e^{at} \\ {c}^{-1}\left\{\frac{n!}{{(s-a)}^{n+1}}\right\}=e^{at}{t}^n \end{gathered}$$

Solve for the solution of differential equation as:

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{Y\right\} \\ =3{\mathcal{L}}^{-1}\left\{\frac{1}{{(s-1)}^2}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{s+6}\right\} \\ =3t{e}^t-e^{-6t} \end{gathered}$$

Since , $y(t-1)=z\left(t\right)$, , replace $t$ by $t-1$

$$\begin{gathered} y\left(t-1\right)=z\left(t\right) \\ =3\left(t-1\right)e^{t-1}-e^{-6\left(t-1\right)} \end{gathered}$$

Therefore, the Initial value for $z\text{'}\text{'}+5z\text{'}-6z=21e^{t-1}$is $z\left(t\right)=3\left(t-1\right)e^{t-1}-e^{-6\left(t-1\right)}$