Q4.3-13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution $\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{26}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+26\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left({\mathrm{c}}_1\cos \left(5\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(5\mathrm{t}\right)\right)$.

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 roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+26\mathrm{y}=0.$

 

Then the auxiliary equation is ${\mathrm{r}}^2-2\mathrm{r}+26=0.$

 

Find the roots of the auxiliary equation.

$\begin{array}{c} \mathrm{r}=\frac{2\pm \sqrt{2^2-4\times 1\times 26}}{2\times 1}\\ \mathrm{r}=\frac{2\pm \sqrt{4-104}}{2}\\ \mathrm{r}=\frac{2\pm \sqrt{-100}}{2}\\ \mathrm{r}=\frac{2\pm 10\mathrm{i}}{2}\\ \mathrm{r}=1\pm 5\mathrm{i}\end{array}$

Step 3: Final answer.

Therefore, the general solution is:

$\begin{array}{c} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{1\times \mathrm{t}}\left({\mathrm{c}}_1\cos \left(5\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(5\mathrm{t}\right)\right)\\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left({\mathrm{c}}_1\cos \left(5\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(5\mathrm{t}\right)\right)\end{array}$