Q21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Devise a modification of the method for Cauchy-Euler equations to find a general solution to the given equation. ${\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{)}}^2\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{7}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{t}\mathbf{>}\mathbf{2}$

Step-by-Step Solution

Answer

The general solution of the given equation  ${(\mathrm{t}-2)}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-7(\mathrm{t}-2)\mathrm{y}\text{'}\left(\mathrm{t}\right)+7\mathrm{y}\left(\mathrm{t}\right)=0,\mathrm{t}>2$ is $\mathrm{y}={\mathrm{c}}_1(\mathrm{t}-2)+{\mathrm{c}}_2{(\mathrm{t}-2)}^7$.

Step 1: Substitute the values.

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

 

Let  $\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{u}\Rightarrow \mathrm{dt}=\mathrm{du}$

 

Therefore, the equation becomes:

 

${\mathrm{u}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{u}\right)-7\mathrm{uy}\text{'}\left(\mathrm{u}\right)+7\mathrm{y}\left(\mathrm{u}\right)=0$

 

Assume  $\mathrm{y}={\mathrm{u}}^{\mathrm{r}}$, then

 

$\begin{array}{l}\mathrm{y}\text{'}={\mathrm{ru}}^{\mathrm{r}-1}\\ \mathrm{y}\text{'}\text{'}=\mathrm{r}(\mathrm{r}-1){\mathrm{u}}^{\mathrm{r}-2}\end{array}$

 

Substitute these equations in the differential equation;

 

$\begin{array}{c}{\mathrm{u}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{u}}^{\mathrm{r}-2}-7{\mathrm{uru}}^{\mathrm{r}-1}+7{\mathrm{u}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)-7\mathrm{r}+7){\mathrm{u}}^{\mathrm{r}}=0\\ {\mathrm{r}}^2-8\mathrm{r}+7=0\end{array}$

Step 2: Finding the roots of the auxiliary equation.

Find the roots of this equation.

$\begin{array}{l}\mathrm{r}=\frac{8\pm \sqrt{8^2-4\times 7\times 1}}{2\times 1}\\ \mathrm{r}=\frac{8\pm \sqrt{64-28}}{2}\\ \mathrm{r}=\frac{8\pm \sqrt{36}}{2}\\ \mathrm{r}=\frac{8\pm 6}{2}\\ r=1,7\end{array}$

Therefore, the general solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{u}}^1+{\mathrm{c}}_2{\mathrm{u}}^7$

 

Substitute $\mathrm{u}=\mathrm{t}-2$ in the above solution, we get:

 

$\mathrm{y}={\mathrm{c}}_1(\mathrm{t}-2)+{\mathrm{c}}_2{(\mathrm{t}-2)}^7$

 

Thus, the solution is $\mathrm{y}={\mathrm{c}}_1(\mathrm{t}-2)+{\mathrm{c}}_2{(\mathrm{t}-2)}^7.$