Q4.3-18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution. $\mathbf{2}\mathbf{y}"\mathbf{+}\mathbf{13}\mathbf{y}\text{'}\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $2\mathrm{y}"+13\mathrm{y}\text{'}-7\mathrm{y}=0$ is  $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\frac{1}{2}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-7\mathrm{t}}$.

Step 1: Differentiate the value of y.

Given differential equation is $2\mathrm{y}"+13\mathrm{y}\text{'}-7\mathrm{y}=0.$

Let  $\mathrm{y}={\mathrm{e}}^{\mathrm{it}}.$

 

Therefore,

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

Step 2: Finding the roots of the auxiliary equation.

Then the auxiliary equation is  $2{\mathrm{r}}^2+13\mathrm{r}-7=0.$

Solve the equation to obtain the roots of the auxiliary equation.

$$\begin{gathered} \mathrm{r}=\frac{-13\pm \sqrt{{13}^2-4\times 2\times -7}}{2\times 2} \\ \mathrm{r}=\frac{-13\pm \sqrt{169+56}}{4} \\ \mathrm{r}=\frac{-13\pm \sqrt{225}}{4} \\ \mathrm{r}=\frac{-13\pm 15}{4} \\ \mathrm{r}=\frac{1}{2}\text{\hspace{0.17em}or\hspace{0.17em} }\mathrm{r}=-7 \end{gathered}$$

Step 3: Final answer.

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\frac{1}{2}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-7\mathrm{t}}$ .