Q4.3-7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

Step-by-Step Solution

Answer

The auxiliary equation for the given differential equation $4\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+6\mathrm{y}=0$ has complex roots and its general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\left({\mathrm{c}}_1\cos \left(\frac{\sqrt{5}\mathrm{t}}{2}\right)\mathrm{t}+{\mathrm{c}}_2\sin \left(\frac{\sqrt{5}\mathrm{t}}{2}\right)\right)$ .

Step 1: Complex conjugate roots.

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$.

Step 2: Finding the roots of the auxiliary equation.

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

 

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

 

The roots of the auxiliary equation are:

$$\begin{gathered} \mathrm{r}=\frac{-4\pm \sqrt{4^2-4\times 4\times 6}}{2\times 4} \\ \mathrm{r}=\frac{-4\pm \sqrt{16-96}}{} \\ \mathrm{r}=\frac{-4\pm 4\mathrm{i}\sqrt{5}}{} \\ \mathrm{r}=-\frac{1}{2}\pm \frac{\sqrt{5}\mathrm{i}}{2} \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is:

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{1}{2}\times \mathrm{t}} \left({\mathrm{c}}_1\cos \left(\frac{\sqrt{5}\mathrm{t}}{2}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\frac{\sqrt{5}\mathrm{t}}{2}\right)\right) \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{1}{2}\mathrm{t}} \left({\mathrm{c}}_1\cos \left(\frac{\sqrt{5}\mathrm{t}}{2}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\frac{\sqrt{5}\mathrm{t}}{2}\right)\right) \end{gathered}$$