17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$17.y^{\text{'}\text{'}}+y^{\text{'}}-y=t^3; y\left(0\right)=1, y^{\text{'}}\left(0\right)=0$

Step-by-Step Solution

Answer

The Initial value for $y\text{'}\text{'}+y\text{'}-y=t^3$is $Y=\frac{s^5+s^4+6}{s^4\left(s^2+s-1\right)}$

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

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

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

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\right\}\left(s\right)=s\mathcal{L}\left\{y\right\}\left(s\right)-y\left(0\right) \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}\left(s\right)=s^2\mathcal{L}\left\{y\right\}\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right) \\ C\left\{t^n\right\}\left(s\right)=\frac{n!}{s^{n+1}} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+\mathcal{L}\left\{y\text{'}\right\}-\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{t^3\right\} \end{gathered}$$

Substitute the properties into the equation.

$\left[s^{2}Y-sy\left(0\right)-y\text{'}\left(0\right)\right]+[sY-y(0\left)\right]-Y=\frac{3!}{s^4}$

Substitute the initial conditions 

$$\begin{gathered} y\left(0\right)=1 and y\text{'}\left(0\right)=0 \\ \left(s^{2}Y-s\right)+\left(sY-1\right)-Y=\frac{6}{s^4} \end{gathered}$$

Isolate the Y variable and solve:

$$\begin{gathered} s^{2}Y+sY-Y=\frac{6}{s^4}+s+1 \\ Y\left(s^2+s-1\right)=\frac{s^5+s^4+6}{s^4} \\ Y=\frac{s^5+s^4+6}{s^4\left(s^2+s-1\right)} \end{gathered}$$

Therefore, the initial value for $y\text{'}\text{'}+y\text{'}-y=t^3$is $Y=\frac{s^5+s^4+6}{s^4\left(s^2+s-1\right)}$