Q4.3-2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

Step-by-Step Solution

Answer

The auxiliary equation for the given differential equation $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$ has complex roots and its general solution is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1\mathbf{cost}\mathbf{+}{\mathbf{c}}_2\mathbf{sint}$.

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $\mathrm{\alpha }\pm \mathrm{i\beta }$, 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.

The auxiliary equation for $\mathrm{y}\text{'}\text{'}+\mathrm{y}=0$ is ${\mathrm{k}}^2+1=0$ . The solutions of the auxiliary equation are:

$$\begin{gathered} {\mathrm{k}}^2+1=0 \\ {\mathrm{k}}^2=-1 \\ \mathrm{k}=\pm \mathrm{i} \end{gathered}$$

Therefore,   $\mathrm{\alpha }=0,\mathrm{\beta }=1$

Step 3: Final answer.

Therefore, the general solution is:

$\mathrm{y}={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}$.