91Ó°ÊÓ

If the auxiliary equation has complex conjugate roots $\mathrm{Î\pm }Â\pm i\mathrm{β}$ , then the general solution is given as:

$y\left(t\right)=c_1{e}^{\mathrm{Î\pm }t}cos\mathrm{β}t+c_2{e}^{\mathrm{Î\pm }t}sin\mathrm{β}t$.

Given differential equation is$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+17\mathrm{y}=0$

Then the auxiliary equation${\mathrm{r}}^2+2\mathrm{r}+17=0$

Solve the auxiliary equation to obtain the roots.

$\begin{array}{c}\mathrm{r}=\frac{-2Â\pm \sqrt{2^2-4×1×17}}{2×1}\\ \mathrm{r}=\frac{-2Â\pm \sqrt{4-68}}{}\\ \mathrm{r}=\frac{-2Â\pm \sqrt{-64}}{}\\ \mathrm{r}=\frac{-2Â\pm 8\mathrm{i}}{}\\ \mathrm{r}=-1Â\pm 4\mathrm{i}\end{array}$

Therefore, the general solution is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-1×\mathrm{t}}\left({\mathrm{c}}_1\cos \left(4\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(4\mathrm{t}\right)\right)\\ ={\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{c}}_1\cos \left(4\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(4\mathrm{t}\right)\right)\end{array}$

Given initial conditions are$\mathrm{y}\left(0\right)=1$ and $\mathrm{y}\text{'}\left(0\right)=-1.$

role="math" localid="1654841671923" $\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{e}}^{-0}\left({\mathrm{c}}_1\cos (4×0)+{\mathrm{c}}_2\sin (4×0)\right)\\ {\mathrm{c}}_1=1\end{array}$

And

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-{\mathrm{e}}^{-\mathrm{t}}({\mathrm{c}}_1\cos 4\mathrm{t}+{\mathrm{c}}_2\sin 4\mathrm{t})+{\mathrm{e}}^{-\mathrm{t}}(-4{\mathrm{c}}_1\mathrm{sint}+4{\mathrm{c}}_2\mathrm{cost})$

Then,

$\begin{array}{l}\mathrm{y}\text{'}\left(0\right)=-{\mathrm{e}}^{-0}\left({\mathrm{c}}_1\cos (4×0)+{\mathrm{c}}_2\sin (4×0)\right)+{\mathrm{e}}^{-0}(-4{\mathrm{c}}_1\sin (4×0)+4{\mathrm{c}}_2\cos (4×0))\\ -{\mathrm{c}}_1+4{\mathrm{c}}_2=-1\end{array}$

Substitute $c_1$ in the above equation

$\begin{array}{c}-1+{\mathrm{c}}_2=-1\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}{\mathrm{c}}_2=0\end{array}$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left(\cos 4\mathrm{t}\right)$.