91Ó°ÊÓ

• 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.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{∫}_{\mathbf{0}}^{∞}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Define $\mathcal{L}\left\{w\right\}\left(s\right)=W\left(s\right)$

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

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

Substitute the properties into the equation

$\left[s^{2}W-sw\left(0\right)-w\text{'}\left(0\right)\right]+W=\frac{2!}{s^{2+1}}+2·\frac{1}{s}$

Substitute the initial conditions

$$\begin{gathered} w\left(0\right)=1andw\text{'}\left(0\right)=-1 \\ \left(s^{2}W-s+1\right)+W=\frac{2}{s^3}+\frac{2}{s} \end{gathered}$$

Simplify and combine fractions.

$⇒s^{2}W+W-s+1=\frac{2s^2+2}{s^3}$

Isolate the W variable as:

$$\begin{gathered} W\left(s^2+1\right)=\frac{2s^2+2}{s^3}+s-1 \\ ⇒W=\frac{s^4-s^3+2s^2+2}{s^3\left(s^2+1\right)} \end{gathered}$$

Find the partial fraction expansion.

Because S is a repeated factor of $s^3\left(s^2+1\right)$we include $s,s^2,and{s}^3)$

$\frac{s^4-s^3+2s^2+2}{s^3\left(s^2+1\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{Ds+E}{s^2+1}$

Combine fractions to equate the numerators and distribute.

$$\begin{gathered} s^4-s^3+2s^2+2=A{s}^2\left(s^2+1\right)+Bs\left(s^2+1\right)+C\left(s^2+1\right)+\left(Ds+E\right)\left(s^3\right) \\ =A{s}^4+A{s}^2+B{s}^3+Bs+C{s}^2+C+D{s}^4+E{s}^3 \end{gathered}$$

Match terms with the same power and solve for variables

$$\begin{gathered} \left(A+D\right)s^4=s^4⇒A+D=1 \\ \left(B+E\right)s^3=-s^3⇒B+E=-1 \\ \left(A+C\right)s^2=2s^2⇒A+C=2 \\ Bs=0s⇒B=0 \end{gathered}$$

Also,

$$\begin{gathered} C=2 \\ ⇒A=2-C=2-2=0 \\ ⇒E=-1-B=-1-0=-1 \end{gathered}$$

$⇒D=1-A=1-0=1$

Substitute the values of into partial fraction expansion.

$W=\frac{s^4-s^3+2s^2+2}{s^3\left(s^2+1\right)}=\frac{2}{s^3}+\frac{s-1}{s^2+1}$

Using the properties listed below take the inverse Laplace transform of $W\left(s\right)$to obtain the solution $w\left(t\right)$$c^{-1}\left\{\frac{b}{s^2+b^2}\right\}=\sin \left(bt\right)$

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}=t^n \\ {\mathcal{L}}^{-1}\left\{\frac{s}{s^2+b^2}\right\}=\cos \left(bt\right) \\ {c}^{-1}\left\{\frac{b}{s^2+b^2}\right\}=\sin \left(bt\right) \end{gathered}$$

Solve as:

$$\begin{gathered} w\left(t\right)={\mathcal{L}}^{-1}\left\{W\right\} \\ ={\mathcal{L}}^{-1}\left\{\frac{2}{s^3}\right\}+{\mathcal{L}}^{-1}\left\{\frac{s}{s^2+1}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{s^2+1}\right\} \\ =t^2+\cos \left(t\right)-\sin \left(t\right) \end{gathered}$$

Therefore, the initial value$w\text{'}\text{'}+w=t^2+2$is$w\left(t\right)=t^2+\cos \left(t\right)-\sin \left(t\right)$