Q4.3-19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution $\mathbf{y}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{3}\mathbf{y}\text{'}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{y}\text{'}\text{'}+3\mathrm{y}\text{'}-5\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{-\mathrm{t}}\cos \left(2\mathrm{t}\right)+{\mathrm{C}}_3{\mathrm{e}}^{-\mathrm{t}}\sin \left(2\mathrm{t}\right)$.

Step 1: Using rational root theorem.

First, one needs to find the auxiliary equation and solve it. One has ${\mathrm{r}}^3+{\mathrm{r}}^2+3\mathrm{r}-5=0$.

 

You can use the rational root theorem.

The first divisor of   $5$  is $\mathbf{1}$ if  will be one solution of the equation and $\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{)}$   will be a factor.

Indeed, you have $1^3+1^2+3-5=0$

Now you can divide  ${\mathrm{r}}^3+{\mathrm{r}}^2+3\mathrm{r}-5$ by  $\mathrm{r}-1$ to get ${\mathrm{r}}^2+2\mathrm{r}+5$.

Step 2: Finding factors and roots.

The equation can be factored as $(\mathrm{r}-1)({\mathrm{r}}^2+2\mathrm{r}+5)=0.$

Since ${\mathrm{r}}^2+2\mathrm{r}+5=0$

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

The roots of the auxiliary equation are $\mathrm{r}=1,\mathrm{r}=-1+2\mathrm{i}$ and $\mathrm{r}=-1-2\mathrm{i}$

Thus, the general solution of the differential equation is: $y\left(t\right)={\mathrm{C}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{-\mathrm{t}}\cos \left(2\mathrm{t}\right)+{\mathrm{C}}_3{\mathrm{e}}^{-\mathrm{t}}\sin \left(2\mathrm{t}\right)$