Q22E - 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.${\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{)}}^2\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{10}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{14}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{t}\mathbf{>}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The general solution to the given equation ${(\mathrm{t}+1)}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+10(\mathrm{t}+1)\mathrm{y}\text{'}\left(\mathrm{t}\right)+14\mathrm{y}\left(\mathrm{t}\right)=0,\mathrm{t}>-1$ is $\mathrm{y}={\mathrm{c}}_1{(\mathrm{t}+1)}^{-2}+{\mathrm{c}}_2{(\mathrm{t}+1)}^{-7}$.

Step 1: Substitute the values:

Given differential equation is ${(\mathrm{t}+1)}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+10(\mathrm{t}+1)\mathrm{y}\text{'}\left(\mathrm{t}\right)+14\mathrm{y}\left(\mathrm{t}\right)=0$

 

Let  $\mathrm{t}+1=\mathrm{u}\Rightarrow \mathrm{dt}=\mathrm{du}$

 

Therefore, the equation becomes:

 

${\mathrm{u}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{u}\right)+10\mathrm{uy}\text{'}\left(\mathrm{u}\right)+14\mathrm{y}\left(\mathrm{u}\right)=0$

 

Assume  $\mathbf{y}\mathbf{=}{\mathbf{u}}^r$, 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;

 $\begin{array}{c}{\mathrm{u}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{u}}^{\mathrm{r}-2}+10{\mathrm{uru}}^{\mathrm{r}-1}+14{\mathrm{u}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)+10\mathrm{r}+14){\mathrm{u}}^{\mathrm{r}}=0\\ {\mathrm{r}}^2+9\mathrm{r}+14=0\end{array}$

The auxiliary equation is:

 ${\mathrm{r}}^2+9\mathrm{r}+14=0$

Step 2: Finding roots

Find the roots of this equation.

$\begin{array}{c}\mathrm{r}=\frac{-9\pm \sqrt{9^2-4\times 14\times 1}}{2\times 1}\\ \mathrm{r}=\frac{-9\pm \sqrt{81-56}}{2}\\ \mathrm{r}=\frac{-9\pm \sqrt{25}}{2}\\ r=-2,-7\end{array}$

Therefore, the general solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{u}}^{-2}+{\mathrm{c}}_2{\mathrm{u}}^{-7}.$

 

Substitute $\mathrm{u}=\mathrm{t}+1$ in the above solution, we get $\mathrm{y}={\mathrm{c}}_1{(\mathrm{t}+1)}^{-2}+{\mathrm{c}}_2{(\mathrm{t}+1)}^{-7}$.

 

Thus, the general solution is $\mathrm{y}={\mathrm{c}}_1{(\mathrm{t}+1)}^{-2}+{\mathrm{c}}_2{(\mathrm{t}+1)}^{-7}.$