Q4.3-14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution. $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+5\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{c}}_1\cos \left(2\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(2\mathrm{t}\right)\right)$.

Step 1: Complex conjugate roots.

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

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

Step 2: Finding roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+5\mathrm{y}=0.$

 

Then the auxiliary equation is ${\mathrm{r}}^2+2\mathrm{r}+5=0.$

 

The roots of the auxiliary equation are:

$$\begin{gathered} \mathrm{r}=\frac{-2\pm \sqrt{2^2-4\times 1\times 5}}{2\times 1} \\ \mathrm{r}=\frac{-2\pm \sqrt{4-20}}{2} \\ \mathrm{r}=\frac{-2\pm \sqrt{-16}}{2} \\ \mathrm{r}=\frac{-2\pm 4\mathrm{i}}{2} \\ \mathrm{r}=-1\pm 2\mathrm{i} \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is:

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