1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems $\mathbf{1}\mathbf{-}\mathbf{14}$, solve the given initial value problem using the method of Laplace transforms.

$\mathbf{1}·\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{4}$

Step-by-Step Solution

Answer

The Initial value for $y^{\text{'}\text{'}}-2y^{\text{'}}+5y=0; y\left(0\right)=2, y^{\text{'}}\left(0\right)=4$is $y\left(t\right)=2e^t\cos \left(2t\right)+e^t\sin \left(2t\right)$

Step 1: 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 initial value of Laplace transform

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

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

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

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\right\}\left(s\right)=s\mathcal{L}\left\{y\right\}\left(s\right)-y\left(0\right) \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}\left(s\right)=s^2\mathcal{L}\left\{y\right\}\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right) \end{gathered}$$

$\mathcal{L}\left\{y\text{'}\text{'}\right\}-2\mathcal{L}\left\{y\text{'}\right\}+5\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{0\right\}$

Substitute the properties into the equation. 

$\left[s^{2}Y-sy\left(0\right)-y\text{'}\left(0\right)\right]-2\left[sY-y\left(0\right)\right]+5Y=0$

Substitute the initial conditions:

 $y\left(0\right)=2$ and  $y\text{'}\left(0\right)=4$

$\left(s^{2}Y-2s-4\right)-2\left(sY-2\right)+5Y=0$

Distribute and simplify

$\Rightarrow s^{2}Y-2sY+5Y-2s=0$

Isolate the Y variable

$$\begin{gathered} Y\left(s^2-2s+5\right)=2s \\ \Rightarrow Y=\frac{2s}{s^2-2s+5} \end{gathered}$$

Complete the denominator. 

This allows us to use Laplace transform properties.

$Y=\frac{2s}{\left(s^2-2s+1\right)-1+5}$

By completing the square, we can apply the following properties.

${\mathcal{L}}^{-}1\left\{\frac{b}{{\left(s-a\right)}^2+b^2}\right\}=e^{at}\sin \left(bt\right)$

${\mathcal{L}}^{-}1\left\{\frac{(s-a)}{{(s-a)}^2+b^2}\right\}=e^{at}\cos \left(bt\right)$

$$\begin{gathered} Y=\frac{2s}{{(s-1)}^2+4} \\ \Rightarrow Y=\frac{2s}{{(s-1)}^2+2^2} \end{gathered}$$

Since $a=1$and $b=2,$we adjust the numerator to match the formulas

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

Using the properties listed above, take the inverse Laplace transform of   to obtain the solution $y\left(t\right)$

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{Y\right\}=2L^{-1}\left\{\frac{s-1}{{(s-1)}^2+2^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{2}{{(s-1)}^2+2^2}\right\} \\ \Rightarrow y\left(t\right)=2e^t\cos \left(2t\right)+e^t\sin \left(2t\right) \\ Therefore, the Initial value for y\text{'}\text{'}-2y\text{'}+5y=0; y\left(0\right)=2, y\text{'}\left(0\right)=4 is \\ y\left(t\right)=2e^t\cos \left(2t\right)+e^t\sin \left(2t\right) \end{gathered}$$