91Ó°ÊÓ

Given differential equation is${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-4\mathrm{ty}\text{'}\left(\mathrm{t}\right)+4\mathrm{y}\left(\mathrm{t}\right)=0$

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;

$\begin{array}{c}{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}-4{\mathrm{trt}}^{\mathrm{r}-1}+4{\mathrm{t}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)-4\mathrm{r}+4){\mathrm{t}}^{\mathrm{t}}=0\\ {\mathrm{r}}^2-5\mathrm{r}+4=0\end{array}$

The auxiliary equation is ${\mathrm{r}}^2−5\mathrm{r}+4=0$.

Find the roots of this equation:

$\begin{array}{c}\mathrm{r}=\frac{5Â\pm \sqrt{5^2-4×4×1}}{2×1}\\ \mathrm{r}=\frac{5Â\pm \sqrt{25-16}}{2}\\ \mathrm{r}=\frac{5Â\pm \sqrt{9}}{2}\\ \mathrm{r}=\frac{5Â\pm 3}{2}\\ r=4,1\end{array}$

Hence, the general solution is$\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^4+{\mathrm{c}}_2{\mathrm{t}}^1.$

Using the given initial conditions;

$\begin{array}{l}\mathrm{y}\left(1\right)={\mathrm{c}}_1{\left(1\right)}^4+{\mathrm{c}}_2\left(1\right)\\ −2={\mathrm{c}}_1+{\mathrm{c}}_2\\ {\mathrm{c}}_1+{\mathrm{c}}_2=−2\text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰}…\left(1\right)\end{array}$

And we have $\mathrm{y}\text{'}\left(\mathrm{t}\right)=4{\mathrm{c}}_1{\mathrm{t}}^3+{\mathrm{c}}_2$then:

$\begin{array}{l}\mathrm{y}\text{'}\left(1\right)=4{\mathrm{c}}_1{\left(1\right)}^3+{\mathrm{c}}_2\\ 4{\mathrm{c}}_1+{\mathrm{c}}_2=-11\text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰}…\left(2\right)\end{array}$

Subtract (2) from (1), we get:

$\begin{array}{l}3{\mathrm{c}}_1=-9\\ \text{ â¶Ä‰â¶Ä‰â€‰}{\mathrm{c}}_1=-3\end{array}$

Therefore,

${\mathrm{c}}_2=1$

Thus, the solution is$\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{t}}^4\mathbf{+}\mathbf{t}.$