Question

Find a general solution for $\mathbf{t}\mathbf{<}\mathbf{0}$.${\mathbf{t}}^2\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathbf{ty}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The general solution of the given equation${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+3\mathrm{ty}\text{'}\left(\mathrm{t}\right)+5\mathrm{y}\left(\mathrm{t}\right)=0$ is$\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-1}\cos \left(2\mathrm{lnt}\right)+{\mathrm{c}}_2{\mathrm{t}}^{-1}\sin \left(2\mathrm{lnt}\right)$ for $\mathrm{t}<0$.

Step-by-step solution

Substitute the values.

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:

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

Finding the roots of the auxiliary equation.

Find the roots of the auxiliary equation.

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

Write the general solution.

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

${\mathrm{y}}_1\left(\mathrm{t}\right)=(-\mathrm{t}{)}^{\mathrm{\alpha }}\cos \left(\mathrm{\beta ln}\right(-\mathrm{t})\mathrm{and}{\mathrm{y}}_2\left(\mathrm{t}\right)=(-\mathrm{t}{)}^{\mathrm{\alpha }}\sin \left(\mathrm{\beta ln}\right(-\mathrm{t}\left)\right)$

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