Q41E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a second linearly independent solution using reduction of order.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{t}\mathbf{>}\mathbf{0}\mathbf{;} \mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{1}}$.

Step-by-Step Solution

Answer

The second linearly independent solution of the given equation ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}-2\mathrm{ty}\text{'}-4\mathrm{y}=0, \mathrm{t}>0; \mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^{-1}$ is $\mathrm{y}=\frac{1}{5{\mathrm{t}}^2}-\frac{1}{20}+{\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^4\mathrm{lnt}$.

Step 1: Finding a homogeneous solution

Given differential equation is ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}-2\mathrm{ty}\text{'}-4\mathrm{y}={\mathrm{t}}^{-1}$

Let $\mathrm{y}={\mathrm{t}}^{\mathrm{r}}$ 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:

$$\begin{gathered} {\mathrm{t}}^2\left(\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}\right)-2\mathrm{t}\left({\mathrm{rt}}^{\mathrm{r}-1}\right)-4{\mathrm{t}}^{\mathrm{r}}=0 \\ \left({\mathrm{r}}^2-3\mathrm{r}-4\right){\mathrm{t}}^{\mathrm{r}}=0 \\ {\mathrm{r}}^2-3\mathrm{r}-4=0 \\ (\mathrm{r}-4)(\mathrm{r}+1)=0 \end{gathered}$$ 

$\mathrm{r}=-1$ and $\mathrm{r}=4$

So, the homogenous solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^4$

Step 2: Finding v 1 , v 2

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

$\begin{array}{rcl}\mathrm{aW}\left[{\mathrm{y}}_1,{\mathrm{y}}_2\right]& =& {\mathrm{t}}^2\left(\left({\mathrm{t}}^{-1}\left(4{\mathrm{t}}^3\right)-\left(-{\mathrm{t}}^{-2}\left({\mathrm{t}}^4\right)\right)\right.\right.\\ & =& {\mathrm{t}}^2\left(5{\mathrm{t}}^2\right)\\ & =& 5{\mathrm{t}}^4\end{array}$

And

$\begin{array}{rcl}{\mathrm{v}}_1& =& \int -\frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{aW}\left[{\mathrm{y}}_1,{\mathrm{y}}_2\right]}\mathrm{dt}\\ & =& \int -\frac{{\mathrm{t}}^{-2}{\mathrm{t}}^4}{5{\mathrm{t}}^4}\mathrm{dt}\\ & =& \frac{1}{5\mathrm{t}}\end{array}$

 

And

$\begin{array}{rcl}{\mathrm{v}}_2& =& \int \frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{aW}\left[{\mathrm{y}}_1,{\mathrm{y}}_2\right]}\mathrm{dt}\\ & =& \int \frac{{\mathrm{t}}^{-2}{\mathrm{t}}^{-1}}{5{\mathrm{t}}^2}\mathrm{dt}\\ & =& -\frac{1}{20{\mathrm{t}}^4}\end{array}$

 

Hence,

$\begin{array}{rcl}{\mathrm{y}}_{\mathrm{p}}& =& \frac{1}{5\mathrm{t}}\left({\mathrm{t}}^{-1}\right)-\frac{1}{20{\mathrm{t}}^4}\left({\mathrm{t}}^4\right)\\ & =& \frac{1}{5{\mathrm{t}}^2}-\frac{1}{20}\end{array}$ 

Therefore, the total solution is $\mathrm{y}=\frac{1}{5{\mathrm{t}}^2}-\frac{1}{20}+{\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^4\mathrm{lnt}$