18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems $\mathbf{15}-\mathbf{24}$ , solve for $\mathbf{Y}\left(\mathbf{s}\right)$ , the Laplace transform of the solution $y\left(t\right)$  to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{-}{\mathbf{e}}^{\mathbf{t}}\mathbf{;} \mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{3}$

Step-by-Step Solution

Answer

The Initial value for $y\text{'}\text{'}-2y\text{'}-y=e^{2t}-e^t$ is $Y\left(s\right)=\frac{s^3-2s^2-s+3}{\left(s^2-2s-1\right)(s-2)(s-1)}$

Step 1: Determine 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 Laplace transform

Applying the Laplace transform and using its linearity we get

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

Solve for the Laplace transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-2\left[sY\left(s\right)-y\left(0\right)\right]-Y\left(s\right)=\frac{1}{\left(s-2\right)\left(s-1\right)} \\ {s}^{2}Y\left(s\right)-s-3-2sY\left(s\right)+2-Y\left(s\right)=\frac{1}{\left(s-2\right)\left(s-1\right)} \end{gathered}$$

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

Solve further as:

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

Therefore, the Initial value for $y\text{'}\text{'}-2y\text{'}-y=e^{2t}-e^t$ is $Y\left(s\right)=\frac{s^3-2s^2-s+3}{\left(s^2-2s-1\right)\left(s-2\right)\left(s-1\right)}$