Question

Solve the given third-order initial value problem forusing the method of Laplace transforms.

$$\begin{gathered} y\text{'}\text{'}\text{'}+y\text{'}\text{'}+3y\text{'}-5y=16e^{-t} \\ y\left(0\right)=0,y\text{'}\left(0\right)=2,y\text{'}\text{'}\left(0\right)=-4 \end{gathered}$$

Short answer

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

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

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 we get

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}\text{'}+y\text{'}\text{'}+3y\text{'}-5y\right\}=\mathcal{L}\left\{16e^{-t}\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\text{'}\right\}+\mathcal{L}\left\{y\text{'}\text{'}\right\}+3\mathcal{L}\left\{y\text{'}\right\}-5\mathcal{L}\left\{y\right\}=-\frac{16}{s+1} \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} \left\{\left[s^{3}Y\left(s\right)-s^{2}y\left(0\right)-sy\text{'}\left(0\right)-y\text{'}\text{'}\left(0\right)\right]+4\left[s^{2}Y\left(s\right)-sy\left(0\right)-y\left(0\right)\right] \\ +\left[sY\left(s\right)-y\left(0\right)\right]-5Y\left(s\right) \\ \right\}=-\frac{16}{s+1} \end{gathered}$$

$$\begin{gathered} s^{3}Y\left(s\right)-2s+4+s^{2}Y\left(s\right)-2+3sY\left(s\right)-5Y\left(s\right)=-\frac{16}{s+1} \\ \left(s^3+s^2+3s-5\right)Y\left(s\right)=-\frac{16}{s+1}+2s-2 \\ Y\left(s\right)=\frac{2s^2+14}{\left(s+1\right)\left(s^3+s^2+3s-5\right)} \end{gathered}$$

Take the partial fraction and solve as:

$$\begin{gathered} \frac{2s^2+14}{\left(s+1\right)\left(s^3+s^2+3s-5\right)}=\frac{2s^2+14}{\left(s+1\right)\left(s-1\right)\left({(s+1)}^2+4\right)} \\ =\frac{A}{s+1}+\frac{B}{s-1}+\frac{Cs+D}{{\left(s+1\right)}^2+4} \end{gathered}$$

Solve further as:

$$\begin{gathered} 2s^2+14=\left\{A\left(s-1\right)\left({\left(s+1\right)}^2+4\right) \\ +B\left(s+1\right)\left({\left(s+1\right)}^2+4\right)+\left(Cs+D\right)\left(s-1\right)\left(s+1\right)\right\} \end{gathered}$$

Using $s=1,-1,0,2$respectively solve as:

$$\begin{gathered} s=1:16=16B\Rightarrow B=1 \\ s=-1:16=-8A\Rightarrow A=-2 \\ s=0:14=15-D\Rightarrow D=1 \\ s=2:22=16+6C\Rightarrow C=1 \end{gathered}$$

Derive the differential equation using inverse Laplace transformation

Therefore

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

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

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

Therefore, the solution for the initial value problem is$y\left(t\right)=-2e^{-t}+e^t+e^{-t}\cos 2t$