Q4.3-9E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}-8\mathrm{y}\text{'}+7\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{7\mathrm{t}}$.

Step 1: Differentiate the value of y.

Given differential equation is $\mathrm{y}\text{'}\text{'}-8\mathrm{y}\text{'}+7\mathrm{y}=0$

Let   $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$

Therefore,

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

Step 2: Finding the roots of the auxiliary equation.

Then the auxiliary equation is  ${\mathrm{r}}^2-8\mathrm{r}+7=0.$

 

Now find the roots of the auxiliary equation.

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{r}}^2\mathbf{-}\mathbf{8}\mathbf{r}\mathbf{+}\mathbf{7}\mathbf{=}\mathbf{0} \\ {\mathbf{r}}^2\mathbf{-}\mathbf{r}\mathbf{-}\mathbf{7}\mathbf{r}\mathbf{+}\mathbf{7}\mathbf{=}\mathbf{0} \\ \mathbf{r}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{-}\mathbf{7}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{7}\mathbf{)}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{7}\mathbf{)}\mathbf{=}\mathbf{0} \end{gathered}$$

                              $\mathbf{r}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{r}\mathbf{=}\mathbf{7}$

Step 3: Final answer.

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{7\mathrm{t}}$