Q4.3-5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{w}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{w}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The auxiliary equation for the given differential equation $\mathrm{w}\text{'}\text{'}+4\mathrm{w}\text{'}+6\mathrm{w}=0$ has complex roots and its general solution is $\mathrm{w}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{c}}_1\cos (\sqrt{2}\mathrm{t})+{\mathrm{c}}_2\sin (\sqrt{2}\mathrm{t})\right)$.

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $\alpha \pm i\beta$ , then the general solution is given as:

$y\left(t\right)=c_1{e}^{\alpha t}cos\beta t+c_2{e}^{\alpha t}sin\beta t$.

Step 2: Finding the roots of the auxiliary equation.

The differential equation is $\mathrm{w}\text{'}\text{'}+4\mathrm{w}\text{'}+6\mathrm{w}=0.$

 

Then the auxiliary equation is ${\mathrm{r}}^2+4\mathrm{r}+6=0.$

$$\begin{gathered} \mathrm{r}=\frac{-4\pm \sqrt{4^2-4\times 1\times 6}}{2\times 1} \\ \mathrm{r}=\frac{-4\pm \sqrt{16-24}}{} \\ \mathrm{r}=\frac{-4\pm 2\mathrm{i}\sqrt{2}}{} \\ \mathrm{r}=-2\pm \sqrt{2}\mathrm{i} \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is:

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