Question

In Problems 29 - 32, use the method of Laplace transforms to find a general solution to the given differential equation by assuming $y\left(0\right)=aandy\text{'}\left(0\right)=b$where a and b are arbitrary constants.

$y\text{'}\text{'}+2y\text{'}+2y=5$

Short answer

The General solution to the given differential equation is

$y\left(t\right)=\frac{5}{2}+\frac{2a-5}{2}{e}^{-t}\cos t+\frac{2a+2b-5}{2}{e}^{-t}\sin t$

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

Determine the solution of the initial value problem:

$$\begin{gathered} \frac{5+a{s}^2+\left(2a+b\right)s}{s\left(s^2+2s+2\right)}=\frac{5+a{s}^2+\left(2a+b\right)s}{s\left({(s+1)}^2+1\right)} \\ =\frac{A}{s}+\frac{B\left(s+1\right)+C}{{\left(s+1\right)}^2+1} \end{gathered}$$Applying the Laplace transform and using its linearity we get

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}+2y\text{'}+2y\right\}=\mathcal{L}\left\{5\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+2\mathcal{L}\left\{y\text{'}\right\}+2\mathcal{L}\left\{y\right\}=\frac{5}{s} \end{gathered}$$

Solve for the partial fraction as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\left(0\right)\right]+2\left[sY\left(s\right)-y\left(0\right)\right]+2Y\left(s\right)=\frac{5}{s} \\ {s}^{2}Y\left(s\right)-as-b+2sY\left(s\right)-2a+2Y\left(s\right)=\frac{5}{s} \\ {s}^{2}Y\left(s\right)+2sY\left(s\right)+2=\frac{5}{s}+as+2a+b \end{gathered}$$

Solve further as:

$$\begin{gathered} \left(s^2+2s+2\right)Y\left(s\right)=\frac{5+a{s}^2+\left(2a+b\right)s}{s} \\ Y\left(s\right)=\frac{5+a{s}^2+\left(2a+b\right)s}{s\left(s^2+2s+2\right)} \end{gathered}$$

Using partial fractions solve as:

$$\begin{gathered} \frac{5+a{s}^2+\left(2a+b\right)s}{s\left(s^2+2s+2\right)}=\frac{5+a{s}^2+\left(2a+b\right)s}{s\left({(s+1)}^2+1\right)} \\ =\frac{A}{s}+\frac{B\left(s+1\right)+C}{{\left(s+1\right)}^2+1} \end{gathered}$$

Resolve the partial fraction as:

$5+a{s}^2+\left(2a+b\right)s=A\left({(s+1)}^2+1\right)+\left(B\left(s+1\right)+C\right)s$

Using $s=0,-1,1$solve for the variables as:

$$\begin{gathered} s=0:5=2A\Rightarrow A=\frac{5}{2} \\ s=-1:5+a-2a-b=A-C\Rightarrow C=\frac{2a+2b-5}{2} \\ s=1:5+2a=2B+10\Rightarrow B=\frac{2a-5}{2} \end{gathered}$$

Use Inverse Laplace transform:

Substitute the values and rewrite as:

$Y\left(s\right)=\frac{5}{2s}+\frac{\left(2a-5\right)\left(s+1\right)+2a+2b-5}{2\left({\left(s+1\right)}^2+1\right)}$

Using the inverse Laplace transform, Obtain the solution of given differential equation:

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{5}{2s}+\frac{(2a-5)(s+1)+2a+2b-5}{2\left({(s+1)}^2+1\right)}\right\}\left(t\right) \\ =\frac{5}{2}\mathcal{L}\left\{\frac{1}{s}\right\}+\frac{2a-5}{2}\mathcal{L}\left\{\frac{s+1}{{(s+1)}^2+1}\right\}+\frac{2a+2b-5}{2}{\mathcal{L}}^{-1}\left\{\frac{1}{{(s+1)}^2+1}\right\} \\ =\frac{5}{2}+\frac{2a-5}{2}{e}^{-t}\cos t+\frac{2a+2b-5}{2}{e}^{-t}\sin t \end{gathered}$$

Therefore, the solution for the differential equation as:

$y\left(t\right)=\frac{5}{2}+\frac{2a-5}{2}{e}^{-t}\cos t+\frac{2a+2b-5}{2}{e}^{-t}\sin t$