Q23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given initial value problem. $\mathbf{w}\text{'}\text{'}\mathbf{-}\mathbf{4}\mathbf{w}\text{'}\mathbf{+}\mathbf{2}\mathbf{w}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{w}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}$$\mathbf{w}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

The solution of the given initial value $\mathrm{w}\text{'}\text{'}-4\mathrm{w}\text{'}+2\mathrm{w}=0$  is $\mathrm{w}\left(\mathrm{t}\right)=\frac{}{2\sqrt{2}}{\mathrm{e}}^{(2+\sqrt{2})\mathrm{t}}-\frac{}{2\sqrt{2}}{\mathrm{e}}^{(2-\sqrt{2})\mathrm{t}}$   when $\mathrm{w}\left(0\right)=0$  and $\mathrm{w}\text{'}\left(0\right)=1$ .

Step 1: Differentiate the value of y.

Given differential equation is $\mathrm{w}\text{'}\text{'}-4\mathrm{w}\text{'}+2\mathrm{w}=0$

 

Let  $\mathrm{w}={\mathrm{e}}^{\mathrm{rt}}$

 

Therefore,

 $\begin{array}{c}\mathrm{w}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}\\ \mathrm{w}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}^2{\mathrm{e}}^{\mathrm{rt}}\end{array}$

 

Step 2: Finding the general solution.

Then the auxiliary equation is ${\mathrm{r}}^2-4\mathrm{r}+2=0$

 

Solve the auxiliary equation to obtain the roots.

$\begin{array}{c}\mathrm{r}=\frac{4\pm \sqrt{4^2-4\times 1\times 2}}{2\times 1}\\ \mathrm{r}=\frac{4\pm \sqrt{16-8}}{}\\ \mathrm{r}=\frac{4\pm \sqrt{8}}{}\\ \mathrm{r}=\frac{4\pm 2\sqrt{2}}{}\\ \mathrm{r}=2\pm \sqrt{2}\end{array}$

Therefore, the general solution is $\mathrm{w}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{(2+\sqrt{2})\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{(2-\sqrt{2})\mathrm{t}}$

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

Given initial conditions are $\mathrm{w}\left(0\right)=0$ and  $\mathrm{w}\text{'}\left(0\right)=1$

 

 $\begin{array}{l} \mathrm{w}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^{(2+\sqrt{2})\times 0}+{\mathrm{c}}_2{\mathrm{e}}^{(2-\sqrt{2})\times 0}\\ {\mathrm{c}}_1+{\mathrm{c}}_2=0\end{array}$

 

And

 $\mathrm{w}\text{'}\left(\mathrm{t}\right)=(2+\sqrt{2}\left){\mathrm{c}}_1{\mathrm{e}}^{(2+\sqrt{2})\mathrm{t}}+(2-\sqrt{2}\right){\mathrm{c}}_2{\mathrm{e}}^{(2-\sqrt{2})\mathrm{t}}$

 

Then $\mathrm{w}\text{'}\left(0\right)=(2+\sqrt{2}\left){\mathrm{c}}_1{\mathrm{e}}^{(2+\sqrt{2})\times 0}+(2-\sqrt{2}\right){\mathrm{c}}_2{\mathrm{e}}^{(2-\sqrt{2})\times 0}$

 

 $\begin{array}{c}2({\mathrm{c}}_1+{\mathrm{c}}_2)+\sqrt{2}({\mathrm{c}}_1-{\mathrm{c}}_2)=1\\ \sqrt{2}({\mathrm{c}}_1+{\mathrm{c}}_1)=1\\ 2\sqrt{2}{\mathrm{c}}_1=1\\ {\mathrm{c}}_1=\frac{}{2\sqrt{2}}\end{array}$

 

Substitute ${\mathrm{c}}_1$  in  $\left(1\right)$$\left(1\right)$

 $\begin{array}{c}\frac{}{2\sqrt{2}}+{\mathrm{c}}_2=0\\ {\mathrm{c}}_2=-\frac{}{2\sqrt{2}}\end{array}$

 

 

Therefore, the solution is  $\mathrm{w}\left(\mathrm{t}\right)=\frac{}{2\sqrt{2}}{\mathrm{e}}^{(2+\sqrt{2})\mathrm{t}}-\frac{}{2\sqrt{2}}{\mathrm{e}}^{(2-\sqrt{2})\mathrm{t}}$.