19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

Step-by-Step Solution

Answer

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

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 initial 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

$$\begin{gathered} \mathcal{L}y\text{'}\text{'}+3y=\mathcal{L}\left\{t^3\right\} \\ \left.\mathcal{L}\left\{y\text{'}\text{'}\right\}+3\mathcal{L}\right\}=\frac{3!}{s^4} \end{gathered}$$

Solve for the Laplace transform as:

$$\begin{gathered} \left[s^{2}Y\left(0\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+3Y\left(s\right)=\frac{6}{s^4} \\ {s}^{2}Y\left(s\right)+3Y\left(s\right)=\frac{6}{s^4} \\ \left(s^2+3\right)Y\left(s\right)=\frac{6}{s^4} \\ Y\left(s\right)=\frac{6}{s^4\left(s^2+3\right)} \end{gathered}$$

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