Question

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

$$\begin{gathered} y\text{'}\text{'}\text{'}-y\text{'}\text{'}+y\text{'}-y=0 \\ y\left(\mathbf{0}\right)=1,y\text{'}\left(\mathbf{0}\right)=1,y\text{'}\text{'}\left(0\right)=3 \end{gathered}$$

Short answer

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

$y\left(t\right)=2e^t-\cos t-\sin t$

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 as follows:

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

Solve for the Laplace transform as:

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

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

Using partial fractions solve as:

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

Solve as:

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

Solve for the variables as:

$\left\{\begin{array}{l}A+B=1\\ C-B=0\\ A-C=3\end{array}\right.\Rightarrow \left\{\begin{array}{l}\begin{array}{l}A=2\\ B=-1\end{array}\\ C=-1\end{array}\right.$

Derive the differential equation using inverse Laplace transformation

Therefore, the simplified equation is:

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

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

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

Therefore

$y\left(t\right)=2e^t-\cos t-\sin t$