Question

Solve the given third-order initial value problem forusing the method of Laplace transforms.$$\begin{gathered} y\text{'}\text{'}\text{'}+3y\text{'}\text{'}+3y\text{'}+y=0 \\ y\left(0\right)=-4,y\text{'}\left(0\right)=4,y\text{'}\text{'}\left(0\right)=-2 \end{gathered}$$

Short answer

The solution for the third-order initial value problem by using Laplace transformation is

$y\left(t\right)=-4e^{-t}+t^2{e}^{-t}$

Step-by-step solution

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

Derive the given equation using Laplace transformation

Applying the Laplace transform and using its linearity solve as:

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}\text{'}+3y\text{'}\text{'}+3y\text{'}+y\right\}=0 \\ \mathcal{L}\left\{y\text{'}\text{'}\text{'}\right\}+3\mathcal{L}\left\{y\text{'}\text{'}\right\}+3\mathcal{L}\left\{y\text{'}\right\}+\mathcal{L}\left\{y\right\}=0 \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} s^{3}Y\left(s\right)+4s^2-4s+2+3s^{2}Y\left(s\right)+12s-12+3sY\left(s\right)+12+Y\left(s\right)=0 \\ \left(s^3+3s^2+3s+1\right)Y\left(s\right)=-4s^2-8s-2 \\ Y\left(s\right)=\frac{-4s^2-8s-2}{s^3+3s^2+3s+1} \end{gathered}$$

Using partial fractions solve as:

$$\begin{gathered} \frac{-4s^2-8s-2}{s^3+3s^2+3s+1}=\frac{-4s^2-8s-2}{{\left(s+1\right)}^3} \\ \frac{-4s^2-8s-2}{s^3+3s^2+3s+1}=\frac{A}{s+1}+\frac{B}{{\left(s+1\right)}^2}+\frac{C}{{\left(s+1\right)}^3} \\ -4s^2-8s-2=A{\left(s+1\right)}^2+B\left(s+1\right)+C \\ -4s^2-8s-2=A{s}^2+\left(2A+B\right)s+A+B+C \end{gathered}$$

From above the solution for the equations are:

$$\begin{gathered} A=-4 \\ 2A+B=-8\Rightarrow \left\{\begin{array}{l}\begin{array}{l}A=-4\\ B=0\end{array}\\ C=-2\end{array}\right. \\ A+B+C=-2 \end{gathered}$$

Derive the differential equation using inverse Laplace transformation

Therefore, solve for the transfer function as:

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

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

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{-\frac{4}{s-1}+\frac{2}{{\left(s+1\right)}^3}\right\}\left(t\right) \\ =-4\mathcal{L}\left\{\frac{1}{s+1}\right\}+2\mathcal{L}\left\{\frac{1}{{\left(s+1\right)}^3}\right\} \\ =-4e^t+t^2{e}^t \end{gathered}$$

Therefore, the equation is$y\left(t\right)=-4e^{-t}+t^2{e}^{-t}$