Q4.3-17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+7\mathrm{y}=0$ is  $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{}{}}\left(({\mathrm{c}}_1\cos \left(\frac{3\sqrt{3}}{2}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\frac{3\sqrt{3}}{2}\mathrm{t}\right)\right)$

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $\alpha \pm i\beta$ , then the general solution is given as: $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$.

Step 2: Finding the roots of the auxiliary equation.

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

 

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

 

The roots of the auxiliary equation are:

$\begin{array}{c} \mathrm{r}=\frac{1\pm \sqrt{1^2-4\times 1\times 7}}{2\times 1}\\ \mathrm{r}=\frac{1\pm \sqrt{1-28}}{2}\\ \mathrm{r}=\frac{1\pm \sqrt{-27}}{2}\\ \mathrm{r}=\frac{1\pm 3\sqrt{3}\mathrm{i}}{2}\end{array}$

Step 3: Final answer.

Therefore, the general solution is:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{1}{2}\mathrm{t}}\left({\mathrm{c}}_1\cos \left(\frac{3\sqrt{3}}{2}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\frac{3\sqrt{3}}{2}\mathrm{t}\right)\right)$