Q4.3-10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{4}\mathbf{y}\text{'}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+8\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{c}}_1\cos \left(2\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(2\mathrm{t}\right)\right)$. 

Step 1: Differentiate the value of y.

Given differential equation is $\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+8\mathrm{y}=0.$

Let $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$

Therefore, we have:

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}$

$\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}^2{\mathrm{e}}^{\mathrm{rt}}$

Step 2: Finding the roots.

Then the auxiliary equation is ${\mathrm{r}}^2+4\mathrm{r}+8=0$

The roots of the auxiliary equation are:

$$\begin{gathered} \mathrm{r}=\frac{-4\pm \sqrt{4^2-4\times 1\times 8}}{2\times 1} \\ \mathrm{r}=\frac{-4\pm \sqrt{16-32}}{2} \\ \mathrm{r}=\frac{-4\pm 4\mathrm{i}}{2} \\ \mathrm{r}=-2\pm 2\mathrm{i} \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is:

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