Q4.3-8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The auxiliary equation for the given differential equation has complex roots. Find a general solution $\mathbf{4}\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{4}\mathbf{y}\text{'}\mathbf{+}\mathbf{26}\mathbf{y}\mathbf{=}\mathbf{0}$. 

Step-by-Step Solution

Answer

The auxiliary equation for the given differential equation $4\mathrm{y}\text{'}\text{'}-4\mathrm{y}\text{'}+26\mathrm{y}=0$ has complex roots and its general solution is  $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{1}{2}\mathrm{t}}\left({\mathrm{c}}_1\cos \left(\frac{5\mathrm{t}}{2}\right)+{\mathrm{c}}_2\sin \left(\frac{5\mathrm{t}}{2}\right)\right)$.

Step 1: Complex conjugate roots.

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

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

Step 2: Finding the roots of the auxiliary equation.

Given differential equation is $4\mathrm{y}\text{'}\text{'}-4\mathrm{y}\text{'}+26\mathrm{y}=0.$

 

Then the auxiliary equation is $4{\mathrm{r}}^2-4\mathrm{r}+26=0$

 

The roots of the auxiliary equation are:

$$\begin{gathered} \mathrm{r}=\frac{4\pm \sqrt{4^2-4\times 4\times 26}}{2\times 4} \\ \mathrm{r}=\frac{4\pm \sqrt{16-416}}{8} \\ \mathrm{r}=\frac{4\pm 20\mathrm{i}}{8} \\ \mathrm{r}=\frac{1}{2}\pm \frac{5\mathrm{i}}{2} \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is:

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