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

Applying the Laplace transform and using its linearity we get

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}+y\text{'}+9y\right\}=0 \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+6\mathcal{L}\left\{y\text{'}\right\}+9\mathcal{L}\left\{y\right\}=0 \\ \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+6\left[sY\left(s\right)-y\left(0\right)\right]+9Y\left(s\right)=0 \end{gathered}$$

Solve for the Laplace transform as:

$$\begin{gathered} s^{2}Y\left(s\right)+s-6+6\left[sY\left(s\right)+1\right]+9Y\left(s\right)=0 \\ {s}^{2}Y\left(s\right)+6sY\left(s\right)+9Y\left(s\right)=-s \\ \left(s^2+6s+9\right)Y\left(s\right)=-s \\ Y\left(s\right)=\frac{-s}{s^2+6s+9} \end{gathered}$$

Using partial fractions solve as:

$$\begin{gathered} \frac{-s}{s^2+6s+9}=\frac{-s}{{\left(s+3\right)}^2}=\frac{A}{s+3}+\frac{B}{{\left(s+3\right)}^2} \\ -s=A\left(s+3\right)+B \end{gathered}$$

Using $s=-3,0$, , respectively, gives

$$\begin{gathered} s=-3:3=B⇒B=3 \\ s=0:0=3A+B⇒A=-1 \end{gathered}$$

Therefore

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

Using the inverse Laplace transform we obtain the solution of given differential equation

$$\begin{gathered} y\left(t\right)=L^{-1}\left\{-\frac{1}{s+3}+\frac{3}{{\left(s+3\right)}^2}\right\}\left(t\right) \\ =-\mathcal{L}\left\{\frac{1}{s+3}\right\}+3\mathcal{L}\left\{\frac{1}{{(s+3)}^2}\right\} \\ =-e^{-3t}+3t{e}^{-3t} \end{gathered}$$

Therefore,

$y\left(t\right)=-e^{-3t}+3t{e}^{-3t}$

Therefore, the initial value for$y\text{'}\text{'}+6y\text{'}+9y=0$is$y\left(t\right)=-e^{-3t}+3t{e}^{-3t}$