Q26E - 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{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,}$$\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$

Step-by-Step Solution

Answer

The solution of the given initial value $\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+\mathrm{y}=0$  is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}-3{\mathrm{te}}^{\mathrm{t}}$ when $\mathrm{y}\left(0\right)=1$ and $\mathrm{y}\text{'}\left(0\right)=-2$  .

 

Step 1: Differentiate the value of y.

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

 

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

 

Therefore,

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

 

Step 2: Finding the general solution.

Then the auxiliary equation is  ${\mathrm{r}}^2-2\mathrm{r}+1=0$

 $\begin{array}{c}{(\mathrm{r}-1)}^2=0\\ \mathrm{r}-1=0\\ \mathrm{r}=1\end{array}$

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{\mathrm{t}}$ .

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

Given initial conditions are $\mathrm{y}\left(0\right)=1$  and   $\mathrm{y}\text{'}\left(0\right)=-2$

 $\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^0+{\mathrm{c}}_2\times 0\times {\mathrm{e}}^0\\ {\mathrm{c}}_1=1\end{array}$

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

 

Then,

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

 

 

Substitute ${\mathbf{c}}_{\mathbf{1}}$  in the above equation

 $\begin{array}{c}1+{\mathrm{c}}_2=-2\\ {\mathrm{c}}_2=-3\end{array}$

 

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}-3{\mathrm{te}}^{\mathrm{t}}$ .