Question

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

Short answer

The general solution of the given equation $\mathrm{u}\text{'}\text{'}+7\mathrm{u}=0$is:

$\mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_1\cos (\sqrt{7}\mathrm{t})+{\mathrm{c}}_2\sin (\sqrt{7}\mathrm{t})\right)$.

Step-by-step solution

Differentiate the value of u.

Given differential equation is$\mathrm{u}\text{'}\text{'}+7\mathrm{u}=0$

Finding roots of the auxiliary equation.

The auxiliary equation is ${\mathrm{r}}^2+7\mathrm{e}=0.$

role="math" localid="1654069093148" $\begin{array}{c}{\mathrm{r}}^2+7=0\\ {\mathrm{r}}^2=-7\\ \mathrm{r}=\pm \sqrt{7}i\end{array}$

Final answer.

Therefore, the general solution is:

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{0\times \mathrm{t}}\left({\mathrm{c}}_1\cos (\sqrt{7}\mathrm{t})+{\mathrm{c}}_2\sin (\sqrt{7}\mathrm{t})\right) \\ \mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_1\cos (\sqrt{7}\mathrm{t})+{\mathrm{c}}_2\sin (\sqrt{7}\mathrm{t})\right) \end{gathered}$$