15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems $\mathbf{15}-\mathbf{24}$, solve for $\mathbf{Y}\left(\mathbf{s}\right)$ , the Laplace transform of the solution  $\mathbf{y}\left(\mathbf{t}\right)$ to the given initial value problem.

$\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{cost}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The Initial value for $y\text{'}\text{'}-3y\text{'}+2y=\cos t$ is $Y\left(s\right)=\frac{-s^2+s-1}{\left(s^2+1\right)\left(s-1\right)\left(s-2\right)}$

Step 1: Determine 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 Laplace transform

Applying the Laplace transform and using its linearity as follows:

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}-3y\text{'}+2y\right\}=\mathcal{L}\left\{\cos t\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}-3\mathcal{L}\left\{y\text{'}\right\}+2\mathcal{L}\left\{y\right\}=\frac{s}{s^2+1} \end{gathered}$$

Solve for the transfer function as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-3\left[sY\left(s\right)-y\left(0\right)\right]+2Y\left(s\right)=\frac{s}{s^2+1} \\ {s}^{2}Y\left(s\right)+1-3sY\left(s\right)+2Y\left(s\right)=\frac{s}{s^2+1} \\ \left(s^2-3s+2\right)Y\left(s\right)=\frac{s}{s^2+1}-1 \\ Y\left(s\right)=\frac{-s^2+s-1}{\left(s^2-3s+2\right)\left(s^2+1\right)} \end{gathered}$$

Since $s^2-3s+2=(s-1)(s-2)$

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

Therefore, the Initial value for $y\text{'}\text{'}-3y\text{'}+2y=\cos t$ is $Y\left(s\right)=\frac{-s^2+s-1}{\left(s^2+1\right)\left(s-1\right)\left(s-2\right)}$