Q38E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find general solutions to the nonhomogeneous Cauchy-Euler equations using a variety of parameters.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{1}}$

Step-by-Step Solution

Answer

The solution of the given equation ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}+3\mathrm{ty}\text{'}+\mathrm{y}={\mathrm{t}}^{-1}$ is $\mathrm{y}=-\frac{{\mathrm{t}}^{-1}\mathrm{lnt}}{2}+\frac{1}{4{\mathrm{t}}^3}+{\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^{-1}\mathrm{lnt}$.

Step 1: Substitute the values

Given differential equation is ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}+3\mathrm{ty}\text{'}+\mathrm{y}={\mathrm{t}}^{-1}$

Let  and then find the solution to the associated homogeneous function,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{t}}^{\mathrm{r}-1} \\ \mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)=\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2} \end{gathered}$$ 

Substitute these in the differential equation:

$\begin{array}{rcl}{\mathrm{t}}^2\left(\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+3\mathrm{t}\left({\mathrm{rt}}^{\mathrm{r}-1}\right)\right)+{\mathrm{t}}^{\mathrm{r}}& =& 0\\ \left({\mathrm{r}}^2+2\mathrm{r}+1\right){\mathrm{t}}^{\mathrm{r}}& =& 0\\ {\mathrm{r}}^2+2\mathrm{r}+1& =& 0\\ {(\mathrm{r}+1)}^2& =& 0\\ \mathrm{r}& =& -1\end{array}$ 

So, the homogenous solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^{-1}\mathrm{lnt}$.

Step 2: Finding v 1

Now find the non-homogenous solution by using the variation of parameter method

$\begin{array}{rcl}\mathrm{aW}\left[{\mathrm{y}}_1,{\mathrm{y}}_2\right]& =& {\mathrm{t}}^2\left(\left({\mathrm{t}}^{-1}\left(-\frac{\mathrm{lnt}-1}{{\mathrm{t}}^2}\right)-\left(-{\mathrm{t}}^{-2}\left({\mathrm{t}}^{-1}\mathrm{lnt}\right)\right)\right.\right.\\ & =& {\mathrm{t}}^2\left(\frac{1}{{\mathrm{t}}^3}\right)\Rightarrow \frac{1}{\mathrm{t}}\end{array}$

And

$\begin{array}{rcl}{\mathrm{v}}_1& =& \int -\frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{aW}\left[{\mathrm{y}}_1,{\mathrm{y}}_2\right]}\mathrm{dt}\\ & =& \int -\frac{{\mathrm{t}}^{-1}{\mathrm{t}}^{-1}\mathrm{lnt}}{\mathrm{t}}\mathrm{dt}\\ & =& \frac{2\mathrm{lnt}+1}{4{\mathrm{t}}^2}\end{array}$

Step 3: Finding v 2

$\begin{array}{rcl}{\mathrm{v}}_2& =& \int \frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{aW}\left[{\mathrm{y}}_1,{\mathrm{y}}_2\right]}\mathrm{dt}\\ & =& \int \frac{{\mathrm{t}}^{-1}{\mathrm{t}}^{-1}}{\mathrm{t}}\mathrm{dt}\\ & =& -\frac{1}{{\mathrm{t}}^2}\end{array}$ 

Therefore,

$\begin{array}{rcl}{\mathrm{y}}_{\mathrm{p}}& =& \frac{2\mathrm{lnt}+1}{4{\mathrm{t}}^2}\left({\mathrm{t}}^{-1}\right)-\frac{1}{{\mathrm{t}}^2}\left({\mathrm{t}}^{-1}\mathrm{lnt}\right)\\ & =& -\frac{{\mathrm{t}}^{-1}\mathrm{lnt}}{2}+\frac{1}{4{\mathrm{t}}^3}\end{array}$ 

 

Therefore, the total solution is $\mathrm{y}=-\frac{{\mathrm{t}}^{-1}\mathrm{lnt}}{2}+\frac{1}{4{\mathrm{t}}^3}+{\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^{-1}\mathrm{lnt}$.