Q4.3-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The auxiliary equation for the given differential equation has complex roots. Find a general solution $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}$. 

Step-by-Step Solution

Answer

The auxiliary equation for the given differential equation $\mathrm{y}\text{'}\text{'}+9\mathrm{y}=0$ has complex roots and its general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos 3\mathrm{t}+{\mathrm{c}}_2\sin 3\mathrm{t}$.

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $\alpha \pm i\beta$, then the general solution is given as:

$y\left(t\right)=c_1{e}^{\alpha t}\cos \beta t+c_2{e}^{\alpha t}\sin \beta t$.

Step 2: Finding the roots of the auxiliary equation.

Given differential equation is $\mathrm{y}"+9\mathrm{y}=0.$

Then the auxiliary equation is $r^2+9=0$.

Finding the roots of the auxiliary equation;

$\begin{array}{c}{\mathrm{r}}^2+9=0\\ {\mathrm{r}}^2=-9\\ \mathrm{r}=\pm 3\mathrm{i}\end{array}$

Step 3: Final answer.

Therefore, the general solution is:

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{0\times \mathrm{t}}\left({\mathrm{c}}_1\cos \left(3\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(3\mathrm{t}\right)\right) \\ ={\mathrm{c}}_1\cos 3\mathrm{t}+{\mathrm{c}}_2\sin 3\mathrm{t} \end{gathered}$$