Q20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given initial value problem for the Cauchy-Euler equation.$\begin{array}{l}{\mathbf{t}}^2\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{7}\mathbf{ty}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\\ \mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{13}\end{array}$

Step-by-Step Solution

Answer

The solution to the given initial value problem

 ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+7\mathrm{ty}\text{'}\left(\mathrm{t}\right)+5\mathrm{y}\left(\mathrm{t}\right)=0;\mathrm{y}\left(1\right)=-1,\mathrm{y}\text{'}\left(1\right)=13$ is  $\mathrm{y}=2{\mathrm{t}}^{-1}-3{\mathrm{t}}^{-5}$

Step 1: Substitute the values.

Given differential equation is ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+7\mathrm{ty}\text{'}\left(\mathrm{t}\right)+5\mathrm{y}\left(\mathrm{t}\right)=0.$

 

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 these equations in the differential equation:

 $\begin{array}{c}{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+7{\mathrm{trt}}^{\mathrm{r}-1}+5{\mathrm{t}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)+7\mathrm{r}+5){\mathrm{t}}^{\mathrm{t}}=0\\ {\mathrm{r}}^2+6\mathrm{r}+5=0\end{array}$

The auxiliary equation is:

 ${\mathrm{r}}^2+6\mathrm{r}+5=0.$

Step 2: Finding the roots of the auxiliary equation.

Find the roots of the auxiliary equation.

$\begin{array}{c}\mathrm{r}=\frac{-6\pm \sqrt{6^2-4\times 5\times 1}}{2\times 1}\\ \mathrm{r}=\frac{-6\pm \sqrt{36-20}}{2}\\ \mathrm{r}=\frac{-6\pm \sqrt{16}}{2}\\ \mathrm{r}=\frac{-6\pm 4}{2}\\ r=-5,-1\end{array}$

Hence, the general solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-1}+{\mathrm{c}}_2{\mathrm{t}}^{-5}.$

Step 3: Finding the values of c 1 , c 2

Using the given initial condition;

 $\begin{array}{l} \mathrm{y}\left(1\right)={\mathrm{c}}_1{\left(1\right)}^{-1}+{\mathrm{c}}_2{\left(1\right)}^{-5}\\ {\mathrm{c}}_1+{\mathrm{c}}_2=-1\dots \left(1\right)\end{array}$

 

And we have $\mathrm{y}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1{\mathrm{t}}^{-2}-5{\mathrm{c}}_2{\mathrm{t}}^{-6}$ then:

 

$\begin{array}{l} \mathrm{y}\text{'}\left(1\right)=-{\mathrm{c}}_1{\left(1\right)}^{-2}-5{\mathrm{c}}_2{\left(1\right)}^{-6}\\ -{\mathrm{c}}_1-5{\mathrm{c}}_2=13 \dots \left(2\right)\end{array}$

 

Add (1) and (2), we get:

 

$\begin{array}{c}-4{\mathrm{c}}_2=12\\ {\mathrm{c}}_2=-3\end{array}$

 

Therefore,

${\mathrm{c}}_1=2$

 

Therefore, the solution is $\mathrm{y}=2{\mathrm{t}}^{-1}-3{\mathrm{t}}^{-5}$.