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

Shift the initial conditions to $t=0$ by defining a new function:

$$\begin{gathered} x\left(t\right)=y\left(t+2\right) \\ x\text{'}\left(t\right)=y\text{'}\left(t+2\right) \\ x\text{'}\text{'}(\left(t\right)=y\text{'}\text{'}\left(t+2\right) \end{gathered}$$

Replace $t$by $t+2$in the condition.

$$\begin{gathered} y\text{'}\text{'}\left(t\right)-y\left(t\right)=t-2 \\ y\text{'}\text{'}\left(t+2\right)-y\left(t+2\right)=\left(t+2\right)-2 \\ y\text{'}\text{'}\left(t+2\right)-y\left(t+2\right)=t \end{gathered}$$

Substitute $x\left(t\right)=y(t+2)$in the equation

$x\left(0\right)=y\left(0+2\right)=y\left(2\right)=3$

$x\text{'}\left(0\right)=y\text{'}\left(0+2\right)=y\text{'}\left(2\right)=0$

$$\begin{gathered} x\text{'}\text{'}\left(t\right)-x\left(t\right)=t \\ x\left(0\right)=3,x\text{'}\left(0\right)=0 \end{gathered}$$

Define $\mathcal{L}\left\{x\right\}\left(s\right)=X\left(s\right)$

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

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

$\mathcal{L}\left\{x\text{'}\text{'}\right\}-\mathcal{L}\left\{x\right\}=\mathcal{L}\left\{t\right\}$

Substitute the properties into the equation.

$\left[s^{2}X-sx\left(0\right)-x\text{'}\left(0\right)\right]-X=\frac{1}{s^2}$

Substitute the initial conditions:

$$\begin{gathered} x\left(0\right)=3andx\text{'}\left(0\right)=0 \\ \left(s^{2}X-3s\right)-X=\frac{1}{s^2} \end{gathered}$$

Isolate the X variable.

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

Find the partial fraction expansion.

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

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

Combine the fractions to equate the numerators.

$$\begin{gathered} 3s^3+1=\left\{As\left(s+1\right)\left(s-1\right)+B\left(s+1\right)\left(s-1\right) \\ +C{s}^2\left(s-1\right)+D{s}^2\left(s+1\right) \\ \right\} \end{gathered}$$

$$\begin{gathered} SolveforvariablesbysettingvaluesofSasfollows: \\ s=0⇒1=A·0-B+C·0+D·0 \\ ⇒B=-1 \\ s=1⇒4=A·0-1·0+C·0+2D \\ ⇒D=2 \\ s=-1⇒-2=A·0-1·0-2C+2·0 \\ ⇒C=1 \\ s=2⇒25=6A-1·3+1·4+2·12 \\ ⇒A=0 \end{gathered}$$

Substitute the values A,B,C,D of into partial fraction expansion

Using the properties listed below take the inverse Laplace transform to obtain the solution $x\left(t\right)$

$$\begin{gathered} c^{-1}\left\{\frac{1}{s-a}\right\}=e^{at} \\ {L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}=t^n \end{gathered}$$

$$\begin{gathered} x\left(t\right)={\mathcal{L}}^{-1}\left\{X\right\} \\ =-{\mathcal{L}}^{-1}\left\{\frac{1}{s^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\right\}+2{\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\} \\ =-t+e^{-t}+2e^t \end{gathered}$$

Since $x(t-2)=y\left(t\right)$replace t by t-2 and solve as:

$$\begin{gathered} x\left(t-2\right)=y\left(t\right) \\ =-\left(t-2\right)+e^{-\left(t-2\right)}+2e^{t-2} \\ =2-t+e^{2-t}+2e^{t-2} \end{gathered}$$

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