16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

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

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 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.

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

$$\begin{gathered} \mathcal{L}\left\{1\right\}\left(s\right)=\frac{1}{s} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+6\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{t^2\right\}-\mathcal{L}\left\{1\right\} \end{gathered}$$

Substitute the properties into the equation 

$\left[s^{2}Y-sy\left(0\right)-y\text{'}\left(0\right)\right]+6Y=\frac{2!}{s^3}-\frac{1}{s}$

Substitute the initial conditions 

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

$\left(s^{2}Y+1\right)+6Y=\frac{2}{s^3}-\frac{1}{s}$

Isolate the Y variable as:

$Y\left(s^2+6\right)=\frac{2}{s^3}-\frac{1}{s}-1=\frac{-s^3-s^2+2}{s^3}$

$Y=\frac{-s^3-s^2+2}{s^3\left(s^2+6\right)}$

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