Q4.3-21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

The solution of the given initial value $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}(2\mathrm{cost}+3\mathrm{sint})$ when $\mathrm{y}\left(0\right)=2$  and $\mathrm{y}\text{'}\left(0\right)=1$.

Step 1: Initial value problem.

An initial value problem is an ordinary differential equation with a given initial condition. The solution of an ordinary differential equation is known as a general solution which consists of an arbitrary constant. The value of an arbitrary constant can be obtained by using the initial condition.

Step 2: Finding the general solution.

Given differential equation is  $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=0.$

Then the auxiliary equation is  ${\mathrm{r}}^2+2\mathrm{r}+2=0.$

$$\begin{gathered} \mathrm{r}=\frac{-2\pm \sqrt{2^2-4\times 1\times 2}}{2\times 1} \\ \mathrm{r}=\frac{-2\pm \sqrt{4-8}}{2} \\ \mathrm{r}=\frac{-2\pm \sqrt{-4}}{2} \\ \mathrm{r}=\frac{-2\pm 2\mathrm{i}}{2} \\ \mathrm{r}=-1\pm \mathrm{i} \end{gathered}$$

Therefore, the general solution is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-1\times \mathrm{t}}\left({\mathrm{c}}_1\cos \left(\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{t}\right)\right)\\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{c}}_1\cos \left(\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{t}\right)\right)\end{array}$

Step 3: Finding the values of c 1 and c 2

Given initial conditions are $\mathrm{y}\left(0\right)=2$ and  $\mathrm{y}\text{'}\left(0\right)=1$

$\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{e}}^{-0}\left({\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)\right)\\ {\mathrm{c}}_1=2\end{array}$

And

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-{\mathrm{e}}^{-\mathrm{t}}({\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint})+{\mathrm{e}}^{-\mathrm{t}}(-{\mathrm{c}}_1\mathrm{sint}+{\mathrm{c}}_2\mathrm{cost})$

Then we have:

$$\begin{gathered} \mathrm{y}\text{'}\left(0\right)=-{\mathrm{e}}^{-0}({\mathrm{c}}_1\cos 0+{\mathrm{c}}_2\sin 0)+{\mathrm{e}}^{-0}(-{\mathrm{c}}_1\sin 0+{\mathrm{c}}_2\cos 0) \\ -{\mathrm{c}}_1+{\mathrm{c}}_2=1 \end{gathered}$$

Substitute ${\mathrm{c}}_1$  in the above equation

$\begin{array}{l}-2+{\mathrm{c}}_2=1\\ {\mathrm{c}}_2=3\end{array}$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}(2\mathrm{cost}+3\mathrm{sint})$.