Question

Find a general solution $\mathbf{z}\mathbf{"}\mathbf{+}\mathbf{10}\mathbf{z}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{z}\mathbf{=}\mathbf{0}$

Short answer

The general solution of the given equation$\mathrm{z}"+10\mathrm{z}\text{'}+25\mathrm{z}=0$is$\mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t}){\mathrm{e}}^{-5\mathrm{t}}$.

Step-by-step solution

Given information.

Given differential equation is$\mathrm{z}"+10\mathrm{z}\text{'}+25\mathrm{z}=0$.

Finding roots of the auxiliary equation.

Then the auxiliary equation is ${\mathrm{r}}^2+10\mathrm{r}+25=0$

Solve the auxiliary equation to obtain the roots.

$$\begin{gathered} {(\mathrm{r}+5)}^2=0 \\ \mathrm{r}+5=0 \\ \mathrm{r}=-5 \end{gathered}$$

Final answer.

Therefore, the general solution is$\mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t}){\mathrm{e}}^{-5\mathrm{t}}$