Q4.3-16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{10}\mathbf{y}\text{'}\mathbf{+}\mathbf{41}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}+10\mathrm{y}\text{'}+41\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-5\mathrm{t}}\left({\mathrm{c}}_1\cos \left(4\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(4\mathrm{t}\right)\right)$.

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:

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

Step 2: Finding the roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}+10\mathrm{y}\text{'}+41\mathrm{y}=0.$

 

Then the auxiliary equation is  ${\mathrm{r}}^2+10\mathrm{r}+41=0.$

 

Solve the auxiliary equation to obtain the roots.

$$\begin{gathered} \mathrm{r}=\frac{-10\pm \sqrt{{10}^2-4\times 1\times 41}}{2\times 1} \\ \mathrm{r}=\frac{-10\pm \sqrt{100-164}}{2} \\ \mathrm{r}=\frac{-10\pm \sqrt{-64}}{2} \\ \mathrm{r}=\frac{-10\pm 8\mathrm{i}}{2} \\ \mathrm{r}=-5\pm 4\mathrm{i} \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-5\times \mathrm{t}}\left({\mathrm{c}}_1\cos \left(4\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(4\mathrm{t}\right)\right)\\ ={\mathrm{e}}^{-5\mathrm{t}}\left({\mathrm{c}}_1\cos \left(4\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(4\mathrm{t}\right)\right)\end{array}$