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

Since the initial conditions are given at $t=-1$,shift them to $t=0$so let $z\left(t\right)=w(t-1)$

Solve the initial value problem as:

$z\text{'}\text{'}\left(t\right)-2z\text{'}\left(t\right)+z\left(t\right)=6t-8,z\left(0\right)=w\left(-1\right)=3,z\text{'}\left(0\right)=w\text{'}\left(-1\right)=7$

Applying the Laplace transform and using its linearity we get

$$\begin{gathered} \mathcal{L}\left\{z\text{'}\text{'}-2z\text{'}+z\right\}=\mathcal{L}\left\{6t-8\right\} \\ \mathcal{L}\left\{z\text{'}\text{'}\right\}-2\mathcal{L}\left\{z\text{'}\right\}+\mathcal{L}\left\{z\right\}=\frac{6}{s^2}-\frac{8}{s} \end{gathered}$$

Solve for the Laplace as:

$\left[s^{2}Z\left(s\right)-sz\left(0\right)-z\text{'}\left(0\right)\right]-2\left[sZ\left(s\right)-z\left(0\right)\right]+Z\left(s\right)=\frac{6}{s^2}-\frac{8}{s}$

$$\begin{gathered} s^{2}Z\left(s\right)-3s-7-2sZ\left(s\right)+6+Z\left(s\right)=\frac{6}{s^2}-\frac{8}{s} \\ {s}^{2}Z\left(s\right)-2sZ\left(s\right)+Z\left(s\right)=\frac{6}{s^2}-\frac{8}{s}+3s+1 \\ \left(s^2-2s+1\right)Z\left(s\right)=\frac{3s^3+s^2-8s+6}{s^2} \\ Z\left(s\right)=\frac{3s^3+s^2-8s+6}{s^2{\left(s-1\right)}^2} \end{gathered}$$

Using partial fractions solve as:

$\frac{3s^3+s^2-8s+6}{s^2{\left(s-1\right)}^2}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-1}+\frac{D}{{\left(s-1\right)}^2}$

Solve for the partial fraction equation as:

$3s^3+s^2-8s+6=As{\left(s-1\right)}^2+B{\left(s-1\right)}^2+C{s}^2\left(s-1\right)+D{s}^2$

Solve for the system of equation as:

$$\begin{gathered} \left\{\begin{array}{l}A+C=3 \\ -2A+B-C+D=1 \\ A-2B=-8 \\ B=6\\ \end{array}\right.⇒\left\{\begin{array}{l}\begin{array}{l}A=4\\ B=6\end{array}\\ C=-1\\ D=2\end{array}\right. \end{gathered}$$

Therefore, $Z\left(s\right)=\frac{4}{s}+\frac{6}{s^2}-\frac{1}{s-1}+\frac{2}{{(s-1)}^2}$

Using the inverse Laplace transform we obtain the solution of $IVP\left(1\right)$

$$\begin{gathered} z\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{4}{s}+\frac{6}{s^2}-\frac{1}{s-1}+\frac{2}{{\left(s-1\right)}^2}\right\}\left(t\right) \\ =\mathcal{L}\left\{\frac{4}{s}\right\}+\mathcal{L}\left\{\frac{6}{s^2}\right\}-\mathcal{L}\left\{\frac{1}{s-1}\right\}+\mathcal{L}\left\{\frac{2}{{\left(s-1\right)}^2}\right\} \\ =4+6t-e^t+2t{e}^t \end{gathered}$$

Since $w\left(t\right)=z(t+1)$obtain the solution of given IVP.

$w\left(t\right)=10+6t+e^{t+1}+2t{e}^{t+1}$

Therefore, the initial value for$w\text{'}\text{'}-2w\text{'}+w=6t-2$is$w\left(t\right)=10+6t+e^{t+1}+2t{e}^{t+1}$