Question

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

$y\text{'}\text{'}-y=g\left(\mathbf{t}\right);y\left(\mathbf{0}\right)=1,y\text{'}\left(0\right)=2,$

Where

$g\left(t\right)=\left\{\begin{array}{l}1,t<3\\ t,t>3\end{array}\right.$

Short answer

The solution for the Laplace transformation is

$Y\left(s\right)=\frac{s^3+2s^2+s+2s{e}^{3s}+e^{3s}}{s^2\left(s^2-1\right)}$

Step-by-step solution

Determine the properties of the Laplace transformation.

Using the properties listed below; take the Laplace transform of the equation.

$$\begin{gathered} \mathbf{L}\left\{\mathbf{y}\mathbf{\text{'}}\right\}\left(\mathbf{s}\right)\mathbf{=}\mathbf{sL}\left\{\mathbf{y}\right\}\left(\mathbf{s}\right)\mathbf{-}\mathbf{y}\left(\mathbf{0}\right) \\ \mathbf{L}\left\{\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\right\}\left(\mathbf{s}\right)\mathbf{=}{\mathbf{s}}^{\mathbf{2}}\mathbf{L}\left\{\mathbf{y}\right\}\left(\mathbf{s}\right)\mathbf{-}\mathbf{sy}\left(\mathbf{0}\right)\mathbf{-}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right) \\ \mathbf{L}\left\{\mathbf{cos}\left(\mathbf{bt}\right)\right\}\mathbf{=}\frac{\mathbf{s}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}} \\ \mathbf{L}\left\{\mathbf{sin}\left(\mathbf{bt}\right)\right\}\mathbf{=}\frac{\mathbf{b}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}} \end{gathered}$$

Use initial condition and find the Y variable

Applying the Laplace transform and using its linearity.

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}-y\right\}=\mathcal{L}\left\{g\left(t\right)\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}-\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{g\left(t\right)\right\} \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-Y\left(s\right)=\mathcal{L}\left\{g\left(t\right)\right\} \\ {s}^{2}Y\left(s\right)-s-2-Y\left(s\right)=\mathcal{L}\left\{g\left(t\right)\right\} \\ \left(s^2-1\right)Y\left(s\right)=\mathcal{L}\left\{g\left(t\right)\right\}+s+2 \\ Y\left(s\right)=\frac{\mathcal{L}\left\{g\left(t\right)\right\}+s+2}{s^2-1} \end{gathered}$$

Determine the expression for $\mathcal{L}\left\{g\right(t\left)\right\}$

Using definition of Laplace transform solve as:

$$\begin{gathered} \mathcal{L}\left\{g\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}g\left(t\right)dt \\ ={\int }_0^3{e}^{-st}dt+{\int }_3^{\infty }t{e}^{-st}dt \\ ={\left[-\frac{e^{-st}}{s}\right]}_0^3+\underset{N\to \infty }{\lim }{\int }_3^{N}t{e}^{-st}dt \\ =\frac{e^{-3s}}{s}+\frac{1}{s}+\left|\begin{array}{cc}u=t& e^{-st}dt=dv\\ du=dt& -\frac{e^{-st}}{s}=v\end{array}\right| \end{gathered}$$

Solve further as:

$$\begin{gathered} \mathcal{L}\left\{g\left(t\right)\right\}=-\frac{e^{-3s}}{s}+\frac{1}{s}+\underset{N\to \infty }{\lim }\left({\left[-\frac{t{e}^{-st}}{s}\right]}_3^N+{\int }_3^N\frac{e^{-st}}{s}\right) \\ =\frac{1-e^{3s}}{s}+\underset{N\to \infty }{\lim }\left(-\frac{N}{s{e}^{NS}}+\frac{3e^{-3s}}{s}-\frac{e^{-NS}}{S^2}+\frac{e^{-3s}}{s^2}\right) \\ =\frac{1-e^{-3s}}{s}+\underset{N\to \infty }{\lim }\left(-\frac{N}{s{e}^{NS}}\right)+\frac{3e^{-3s}}{s}-\underset{N\to \infty }{\lim }\frac{e^{NS}}{s^2}+\frac{e^{3s}}{s^2} \\ =\frac{1-e^{-3s}}{s}+\frac{3e^{-3s}}{s}+\frac{e^{-3s}}{s^2} \end{gathered}$$

Solve further as:

$\mathcal{L}\left\{g\left(t\right)\right\}=\frac{s+2s{e}^{-3s}+e^{3s}}{s^2}$

Substituting this into (1) gives

$Y\left(s\right)=\frac{s^3+2s^2+s+2s{e}^{3s}+e^{3s}}{s^2\left(s^2-1\right)}$