Question

In Problems 15-24, 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.

$$\begin{gathered} y\text{'}\text{'}+4y=g\left(t\right);y\left(0\right)=-1,y\text{'}\left(0\right)=0 \\ where \\ g\left(t\right)=\left\{\begin{array}{l}t,t<2\\ 5,t>2\\ \end{array}\right. \end{gathered}$$

Short answer

The solution for the Laplace transformation is

$Y\left(s\right)=\frac{1+3s{e}^{-2s}-s^{-2s}-s^3}{s^2\left(s^2+4\right)}$

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 we get

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}+4y\right\}=\mathcal{L}\left\{g\left(t\right)\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+4\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{g\left(t\right)\right\} \end{gathered}$$

Solve for the Laplace transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+4Y\left(s\right)=\mathcal{L}\left\{g\left(t\right)\right\} \\ {s}^{2}Y\left(s\right)+s+4Y\left(s\right)=\mathcal{L}\left\{g\left(t\right)\right\} \\ \left(s^2+4\right)Y\left(s\right)=\mathcal{L}\left\{g\left(t\right)\right\}-s \\ Y\left(s\right)=\frac{\mathcal{L}\left\{g\left(t\right)\right\}-s}{s^2+4} \end{gathered}$$

Solve for $\mathcal{L}\left\{g\left(t\right)\right\}$is

By the definition of the Laplace transform solve as:

$$\begin{gathered} \mathcal{L}\left\{g\left(t\right)\right\}={\int }_0^{\infty }{e}^{st}g\left(t\right)dt \\ ={\int }_0^2{e}^{st}tdt+{\int }_0^{\infty }5e^{st}dt \\ =\left|\begin{array}{cc}u=t& e^{-st}dt=dv\\ du=dt& -\frac{e^{-st}}{s}=v\end{array}\right|+\underset{N\to \infty }{\lim }{\int }_2^{N}5e^{st}dt \\ =\left({\left[-\frac{t{e}^{-st}}{s}\right]}_0^2+{\int }_0^2\frac{e^{-st}}{s}dt\right)+\underset{N\to \infty }{\lim }{\left[-\frac{5e^{-st}}{s}\right]}_2^N \end{gathered}$$

Solve further as:

$$\begin{gathered} \mathcal{L}\left\{g\left(t\right)\right\}=-\frac{2e^{-2s}}{s}+{\left[-\frac{e^{-st}}{s^2}\right]}_0^2+\underset{N\to \infty }{\lim }\left(-\frac{5e^{-Ns}}{s}+\frac{5e^{2s}}{s}\right) \\ =-\frac{2e^{2s}}{s}-\frac{e^{2s}}{s^2}+\frac{1}{s^2}+\frac{5e^{2s}}{s} \\ =\frac{1+3s{e}^{2s}-e^{2s}}{s^2} \end{gathered}$$

Substituting this into (1) gives:

$Y\left(s\right)=\frac{1+3s{e}^{-2s}-s^{-2s}-s^3}{s^2\left(s^2+4\right)}$