Q4.3-4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{26}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The auxiliary equation for the given differential equation $\mathrm{y}\text{'}\text{'}-10\mathrm{y}\text{'}+26\mathrm{y}=0$  has complex roots and its general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}\left({\mathrm{c}}_1\cos \left(\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{t}\right)\right)$.

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:

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

Step 2: The auxiliary equation.

Assume that $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$ is a solution to the given equation.

Since the given equation is of order two, differentiate $y$ with respect to $x$ twice:

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}} \\ {\mathrm{y}}^{"}\left(\mathrm{t}\right)={\mathrm{r}}^2{\mathrm{e}}^{\mathrm{rt}} \end{gathered}$$

Substitute $\mathrm{y}",\mathrm{y}\text{'}$ and $y$ in the given equation to obtain: ${\mathrm{e}}^{\mathrm{rt}}({\mathrm{r}}^2-10\mathrm{r}+26)=0.$

 

Then the auxiliary equation is ${\mathrm{r}}^2-10\mathrm{r}+26=0.$ 

Step 3: Finding the roots.

Solve for the roots of the auxiliary equation

$$\begin{gathered} {\mathrm{r}}_{1/2}=\frac{10\pm \sqrt{{10}^2-4\times 1\times 26}}{2\times 1} \\ =\frac{10\pm \sqrt{100-104}}{} \\ =\frac{10\pm 2\mathrm{i}}{} \\ =5\pm \mathrm{i} \end{gathered}$$

Step 4: Final answer.

Since it has a complex conjugate of the form $\mathrm{r}=\mathrm{\alpha }\pm \mathrm{i\beta }$ for $\alpha =5$ and $\beta =1$

Thus, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}\left({\mathrm{c}}_1\cos \left(\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{t}\right)\right)$.