Question

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.${\mathbf{t}}^2\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dt}}^2}\mathbf{+}\mathbf{2}\mathbf{t}\frac{\mathrm{dy}}{\mathrm{dt}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The general equation is$\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-3}+{\mathrm{c}}_2{\mathrm{t}}^2$.

Step-by-step solution

Find the auxiliary equation. 

Given differential equation ${\mathrm{t}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+2\mathrm{t}\frac{\mathrm{dy}}{\mathrm{dt}}-6\mathrm{y}=0\dots \left(1\right)$

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 all values in equation (1), and we get:

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

Determine the general equation. 

The roots of the equation are:

$\begin{array}{l}{\mathrm{r}}^2+3\mathrm{r}-2\mathrm{r}-6=0\\ (\mathrm{r}+3)(\mathrm{r}-2)=0\\ \mathrm{r}=-3,2\end{array}$

Thus, the general solution is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{t}}^{-3}\mathbf{+}{\mathbf{c}}_2{\mathbf{t}}^2$.