91Ó°ÊÓ

Given differential equation is${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+3\mathrm{ty}\text{'}\left(\mathrm{t}\right)+5\mathrm{y}\left(\mathrm{t}\right)=0$

Assume $\mathrm{y}={\mathrm{t}}^{\mathrm{r}}$, then

$\begin{array}{l}\mathrm{y}\text{'}={\mathrm{rt}}^{\mathrm{r}-1}\\ \mathrm{y}\text{'}\text{'}=\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}\end{array}$

Substitute these equations in the differential equation;

$\begin{array}{c}{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+3{\mathrm{trt}}^{\mathrm{r}-1}+5{\mathrm{t}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)+3\mathrm{r}+5){\mathrm{t}}^{\mathrm{t}}=0\\ {\mathrm{r}}^2+2\mathrm{r}+5=0\end{array}$

The auxiliary equation is:

${\mathrm{r}}^2+2\mathrm{r}+5=0$

Find the roots of the auxiliary equation.

$\begin{array}{c}\mathrm{r}=\frac{-2Â\pm \sqrt{2^2-4×5×1}}{2×1}\\ \mathrm{r}=\frac{-2Â\pm \sqrt{4-20}}{2}\\ \mathrm{r}=\frac{-2Â\pm \sqrt{-16}}{2}\\ \mathrm{r}=-1Â\pm 2\mathrm{i}\end{array}$

When the roots to the characteristic equation are complex, $\mathrm{r}=\mathrm{Î\pm }Â\pm \mathrm{β¾\pm }$and if$\mathrm{t}>0$ then the linearly independent solutions are${\mathrm{y}}_1\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{Î\pm }}\cos \left(\mathrm{β\pm ô²Ô³Ù}\right)$ and${\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{Î\pm }}\sin \left(\mathrm{β\pm ô²Ô³Ù}\right).$ But if we have that $\mathrm{t}<0$, then the linearly independent solutions are given as:

${\mathrm{y}}_1\left(\mathrm{t}\right)=(-\mathrm{t}{)}^{\mathrm{Î\pm }}\cos \left(\mathrm{β\pm ô²Ô}\right(-\mathrm{t})\text{ â¶Ä‰â¶Ä‰}\mathrm{and}\text{ â¶Ä‰â¶Ä‰}{\mathrm{y}}_2\left(\mathrm{t}\right)=(-\mathrm{t}{)}^{\mathrm{Î\pm }}\sin \left(\mathrm{β\pm ô²Ô}\right(-\mathrm{t}\left)\right)$

Therefore, the general solution is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{t}}^{-1}\mathbf{cos}\left(\mathbf{2}\mathbf{ln}(−\mathbf{t})\right)\mathbf{+}{\mathbf{c}}_2{\mathbf{t}}^{-1}\mathbf{sin}\left(\mathbf{2}\mathbf{ln}(−\mathbf{t})\right).$