Question

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.

${\mathbf{t}}^2\frac{{\mathbf{d}}^2\mathbf{z}}{{\mathbf{dt}}^2}\mathbf{+}\mathbf{5}\mathbf{t}\frac{\mathrm{dz}}{\mathrm{dt}}\mathbf{+}\mathbf{4}\mathbf{z}\mathbf{=}\mathbf{0}$

Short answer

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

Step-by-step solution

Find the auxiliary equation.

Given differential equation${\mathrm{t}}^2\frac{{\mathrm{d}}^2\mathrm{z}}{{\mathrm{dt}}^2}+5\mathrm{t}\frac{\mathrm{dz}}{\mathrm{dt}}+4\mathrm{z}=0\dots \left(1\right)$

Assume $\mathrm{z}={\mathrm{t}}^{\mathrm{r}}$then we have:

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

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

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

Determine the general equation.

The roots of the equation are:

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

Thus, the general equation is $\mathbf{z}\mathbf{=}{\mathbf{c}}_1{\mathbf{t}}^{-2}\mathbf{+}{\mathbf{c}}_2{\mathbf{t}}^{-2}\mathbf{lnt}$.