21E - Fundamentals Of Differential Equations And Boundary Value Problems

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.

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

Step-by-Step Solution

Answer

The solution for the Laplace transformation is

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

Step 1: Derive the given equation using Laplace transformation

Define $\mathcal{L}\left\{y\right\}\left(s\right)=Y\left(s\right)$

$\mathcal{L}\left\{y\text{'}\text{'}\right\}-2\mathcal{L}\left\{y\text{'}\right\}+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{\cos \left(t\right)\right\}-\mathcal{L}\left\{\sin \left(t\right)\right\}$

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

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

Substitute the properties into the equation.

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

Step 2: Use initial condition and find the Y variable

Solve for the Laplace transform as:

$s^{2}Y-s-3-2sY+2+Y=\frac{s-1}{s^2+1}$

Substitute the initial conditions:

$y\left(0\right)=1 and y^{\text{'}}\left(0\right)=3$

Isolate the Y variable.

$$\begin{gathered} s^{2}Y-2sY+Y-s-1=\frac{s-1}{s^2+1} \\ Y\left(s^2-2s+1\right)=\frac{s-1}{s^2+1}+s+1 \\ Y(s-1{)}^2=\frac{s-1+(s+1)\left(s^2+1\right)}{s^2+1} \\ Y=\frac{s^3+s^2+2s}{\left(s^2+1\right){(s-1)}^2} \end{gathered}$$