4E - 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.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{12}{\mathbf{e}}^{\mathbf{t}}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{7}$

Step-by-Step Solution

Answer

The Initial value for $y\text{'}\text{'}+6y\text{'}+5y=12e^t$ is $y\left(t\right)=-e^{-5t}-e^{-t}+e^t$

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

Applying the Laplace transform and using its linearity we get

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}+6y\text{'}+5y\right\}=\mathcal{L}\left\{12e^t\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+6\mathcal{L}\left\{y\text{'}\right\}+5\mathcal{L}\left\{y\right\}=\frac{12}{s-1} \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+6\left[sY\left(s\right)-y\left(0\right)\right]+5Y\left(s\right)=\frac{12}{s-1} \\ {s}^{2}Y\left(s\right)+s-7+6\left[sY\left(s\right)+1\right]+5Y\left(s\right)=\frac{12}{s-1} \end{gathered}$$

$$\begin{gathered} s^{2}Y\left(s\right)+6sY\left(s\right)+5Y\left(s\right)=\frac{12}{s-1}+1-s \\ \left(s^2+6s+5\right)Y\left(s\right)=\frac{-s^2+2s+11}{s-1} \\ Y\left(s\right)=\frac{-s^2+2s+11}{\left(s^2+6s+5\right)\left(s-1\right)} \end{gathered}$$

Using partial fractions solve as:

$$\begin{gathered} \frac{-s^2+2s+11}{\left(s^2+6s+5\right)\left(s-1\right)}=\frac{-s^2+2s+11}{\left(s+5\right)\left(s+1\right)\left(s-1\right)} \\ =\frac{A}{s+5}+\frac{B}{s+1}+\frac{C}{s-1} \end{gathered}$$

Solve further as:

$-s^2+2s+11=A\left(s^2-1\right)+B\left(s+5\right)\left(s-1\right)+C\left(s+5\right)\left(s+1\right)$

Using $s=1,-1,-5$ respectively, gives

$$\begin{gathered} s=1:-1+2+11=12C\Rightarrow C=1 \\ s=-1:-1-2+11=-8B\Rightarrow B=-1 \\ s=-5:-25-10+11=24A\Rightarrow A=-1 \end{gathered}$$

Therefore, the equation becomes:

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

Using the inverse Laplace transform we obtain the solution of given differential equation 

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{-\frac{1}{s+5}-\frac{1}{s+1}+\frac{1}{s-1}\right\}\left(t\right) \\ =-\mathcal{L}\left\{\frac{1}{s+5}\right\}-\mathcal{L}\left\{\frac{1}{s+1}\right\}+\mathcal{L}\left\{\frac{1}{s-1}\right\} \\ =-e^{-5t}-e^{-t}+e^t \end{gathered}$$

Therefore,

$y\left(t\right)=-e^{-5t}-e^{-t}+e^t$

Therefore, the initial value for $y\text{'}\text{'}+6y\text{'}+5y=12e^t$is $y\left(t\right)=-e^{-5t}-e^{-t}+e^t$