Question

In Problems $\mathbf{15}-\mathbf{24}$, 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.

$y\text{'}\text{'}-6y\text{'}+5y=t{e}^t;y\left(0\right)=2,y\text{'}\left(0\right)=-1$

Short answer

The solution for the Laplace transformation is

$Y\left(s\right)=\frac{2s^3-17s^2+28s-12}{\left(s^2-6s+5\right){\left(s-1\right)}^2}$

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

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}-6y\text{'}+5y\right\}=\mathcal{L}\left\{t{e}^t\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}-6\mathcal{L}\left\{y\text{'}\right\}+5\mathcal{L}\left\{y\right\}=\frac{1}{{\left(s-1\right)}^2} \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-6\left[sY\left(s\right)-y\left(0\right)\right]+5Y\left(s\right)=\frac{1}{{\left(s-1\right)}^2} \\ {s}^{2}Y\left(s\right)-2s+1-6sY\left(s\right)+12+5Y\left(s\right)={\left(s-1\right)}^2 \\ {s}^{2}Y\left(s\right)-6sY\left(s\right)+5Y\left(s\right)={\left(s-1\right)}^2+2s-13 \end{gathered}$$

Solve further as:

$$\begin{gathered} \left(s^2-6s+5\right)Y\left(s\right)=\frac{2s^3-17s^2+28s-12}{{\left(s-1\right)}^2} \\ Y\left(s\right)=\frac{2s^3-17s^2+28s-12}{\left(s^2-6s+5\right){\left(s-1\right)}^2} \end{gathered}$$

Therefore, the Laplace transform is$Y\left(s\right)=\frac{2s^3-17s^2+28s-12}{\left(s^2-6s+5\right){\left(s-1\right)}^2}$