Q4.3-20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution.$\mathbf{y}\text{'}\text{'}\text{'}\mathbf{-}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

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

Step 1: Using rational root theorem.

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

The first divisor of   $2$ is $1$  if  $\mathbf{1}$ will be one solution of the equation and $(\mathrm{r}-1)$  will be a factor.

That doesn't happen, but next, you can try with $\mathbf{-}\mathbf{1}$  so that $\mathbf{r}\mathbf{+}\mathbf{1}$  would be a factor.

Step 2: Finding factor

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

Therefore, the equation can be factored as $(\mathrm{r}+1)({\mathrm{r}}^2-2\mathrm{r}+2)=0$

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

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

 

Step 3: Finding roots.

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

Thus, the general solution of the differential equation is:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{\mathrm{t}}\cos \left(\mathrm{t}\right)+{\mathrm{C}}_3{\mathrm{e}}^{\mathrm{t}}\sin \left(\mathrm{t}\right)$