Q25E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{2}\mathbf{y}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\pi }\mathbf{,}$$\mathbf{y}\text{'}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The solution of the given initial value $\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=0$   is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}(-\mathrm{cost}+\mathrm{sint})$   when $\mathrm{y}\left(\mathrm{\pi }\right)={\mathrm{e}}^{\mathrm{\pi }}$  and $\mathrm{y}\text{'}\left(\mathrm{\pi }\right)=0$ .

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $\mathbf{\alpha }\mathbf{\pm }\mathbf{i\beta }$, then the general solution is given as: $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$

 

  

Step 2: Finding the roots of the auxiliary equation

Given differential equation is $\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=0.$

 

Then the auxiliary equation is ${\mathrm{r}}^2-2\mathrm{r}+2=0.$

 

 $\begin{array}{l}\mathrm{r}=\frac{2\pm \sqrt{2^2-4\times 1\times 2}}{2\times 1}\\ \mathrm{r}=\frac{2\pm \sqrt{4-8}}{}\\ \mathrm{r}=\frac{2\pm \sqrt{-4}}{}\\ \mathrm{r}=\frac{2\pm 2\mathrm{i}}{}\\ \mathrm{r}=1\pm \mathrm{i}\end{array}$

 

Therefore, the general solution is:

 

 $\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{1\times \mathrm{t}}\left({\mathrm{c}}_1\cos \left(\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{t}\right)\right)\\ ={\mathrm{e}}^{\mathrm{t}}\left({\mathrm{c}}_1\cos \left(\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{t}\right)\right)\end{array}$

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

Given initial conditions are $\mathrm{y}\left(\mathrm{\pi }\right)={\mathrm{e}}^{\mathrm{\pi }}$ and  $\mathrm{y}\text{'}\left(\mathrm{\pi }\right)=0$

 

 $\begin{array}{l} \mathrm{y}\left(\mathrm{\pi }\right)={\mathrm{e}}^{\mathrm{\pi }}\left({\mathrm{c}}_1\cos \left(\mathrm{\pi }\right)+{\mathrm{c}}_2\sin \left(\mathrm{\pi }\right)\right)\\ -{\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\pi }}={\mathrm{e}}^{\mathrm{\pi }}\\ {\mathrm{c}}_1=-1\end{array}$

And  $\mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}({\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint})+{\mathrm{e}}^{\mathrm{t}}(-{\mathrm{c}}_1\mathrm{sint}+{\mathrm{c}}_2\mathrm{cost})$ 

 

Then  $\mathrm{y}\text{'}\left(\mathrm{\pi }\right)={\mathrm{e}}^{\mathrm{\pi }}\left({\mathrm{c}}_1\cos \left(\mathrm{\pi }\right)+{\mathrm{c}}_2\sin \left(\mathrm{\pi }\right)\right)+{\mathrm{e}}^{\mathrm{\pi }}(-{\mathrm{c}}_1\sin \left(\mathrm{\pi }\right)+{\mathrm{c}}_2\cos \left(\mathrm{\pi }\right))$

 

 $\begin{array}{c}-{\mathrm{e}}^{\mathrm{\pi }}({\mathrm{c}}_1+{\mathrm{c}}_2)=0\\ {\mathrm{c}}_1+{\mathrm{c}}_2=0\\ {\mathrm{c}}_2=-{\mathrm{c}}_1\end{array}$

 

Then  ${\mathrm{c}}_2=1$

 

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}(-\mathrm{cost}+\mathrm{sint})$