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

Question

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{z}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{z}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The auxiliary equation for the given differential equation $\mathrm{z}\text{'}\text{'}-6\mathrm{z}\text{'}+10\mathrm{z}=0$ has complex roots and its general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}({\mathrm{c}}_1\cos 3\mathrm{t}+{\mathrm{c}}_2\sin 3\mathrm{t})$.

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $\mathrm{\alpha }\pm \mathrm{i\beta }$ , 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 $\mathrm{z}\text{'}\text{'}-6\mathrm{z}\text{'}+10\mathrm{z}=0.$ 

 

Then the auxiliary equation is ${\mathrm{r}}^2-6\mathrm{r}+10=0.$

 

The roots of the auxiliary equation are given below:

$$\begin{gathered} \mathrm{r}=\frac{6\pm \sqrt{6^2-4\times 1\times 10}}{2\times 1} \\ \mathrm{r}=\frac{6\pm \sqrt{36-40}}{} \\ \mathrm{r}=\frac{6\pm 2\mathrm{i}}{} \\ \mathrm{r}=3\pm \mathrm{i} \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is:

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{3\times \mathrm{t}}\left({\mathrm{c}}_1\cos \left(\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{t}\right)\right) \\ ={\mathrm{e}}^{\mathrm{t}}({\mathrm{c}}_1\cos 3\mathrm{t}+{\mathrm{c}}_2\sin 3\mathrm{t}) \end{gathered}$$