91Ó°ÊÓ

Given differential equation is${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+7\mathrm{ty}\text{'}\left(\mathrm{t}\right)+5\mathrm{y}\left(\mathrm{t}\right)=0.$

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

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

Substitute these equations in the differential equation:

$\begin{array}{c}{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+7{\mathrm{trt}}^{\mathrm{r}-1}+5{\mathrm{t}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)+7\mathrm{r}+5){\mathrm{t}}^{\mathrm{t}}=0\\ {\mathrm{r}}^2+6\mathrm{r}+5=0\end{array}$

The auxiliary equation is:

${\mathrm{r}}^2+6\mathrm{r}+5=0.$

Find the roots of the auxiliary equation.

$\begin{array}{c}\mathrm{r}=\frac{-6Â\pm \sqrt{6^2-4×5×1}}{2×1}\\ \mathrm{r}=\frac{-6Â\pm \sqrt{36-20}}{2}\\ \mathrm{r}=\frac{-6Â\pm \sqrt{16}}{2}\\ \mathrm{r}=\frac{-6Â\pm 4}{2}\\ r=−5,−1\end{array}$

Hence, the general solution is$\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^{-5}.$

Using the given initial condition;

$\begin{array}{l}\mathrm{y}\left(1\right)={\mathrm{c}}_1{\left(1\right)}^{-1}+{\mathrm{c}}_2{\left(1\right)}^{-5}\\ {\mathrm{c}}_1+{\mathrm{c}}_2=-1\text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â€‰}…\left(1\right)\end{array}$

And we have $\mathrm{y}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1{\mathrm{t}}^{-2}-5{\mathrm{c}}_2{\mathrm{t}}^{-6}$then:

$\begin{array}{l}\mathrm{y}\text{'}\left(1\right)=-{\mathrm{c}}_1{\left(1\right)}^{-2}-5{\mathrm{c}}_2{\left(1\right)}^{-6}\\ -{\mathrm{c}}_1-5{\mathrm{c}}_2=13\text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰}…\left(2\right)\end{array}$

Add (1) and (2), we get:

$\begin{array}{c}-4{\mathrm{c}}_2=12\\ \text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰}{\mathrm{c}}_2=-3\end{array}$

Therefore,

${\mathrm{c}}_1=2$

Therefore, the solution is $\mathrm{y}=2{\mathrm{t}}^{-1}-3{\mathrm{t}}^{-5}$.