Q22E - 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{17}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}$$\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$.    

Step-by-Step Solution

Answer

The solution of the given initial value $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+17\mathrm{y}=0$  is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left(\cos 4\mathrm{t}\right)$   when $\mathrm{y}\left(0\right)=1$  and $\mathrm{y}\text{'}\left(0\right)=-1.$  

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $\alpha \pm i\beta$ , then the general solution is given as:

 

 $y\left(t\right)=c_1{e}^{\alpha t}cos\beta t+c_2{e}^{\alpha t}sin\beta t$.

Step 2: Finding the roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+17\mathrm{y}=0$ 

 

Then the auxiliary equation ${\mathrm{r}}^2+2\mathrm{r}+17=0$

 

Solve the auxiliary equation to obtain the roots.

$\begin{array}{c}\mathrm{r}=\frac{-2\pm \sqrt{2^2-4\times 1\times 17}}{2\times 1}\\ \mathrm{r}=\frac{-2\pm \sqrt{4-68}}{}\\ \mathrm{r}=\frac{-2\pm \sqrt{-64}}{}\\ \mathrm{r}=\frac{-2\pm 8\mathrm{i}}{}\\ \mathrm{r}=-1\pm 4\mathrm{i}\end{array}$

Therefore, the general solution is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-1\times \mathrm{t}}\left({\mathrm{c}}_1\cos \left(4\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(4\mathrm{t}\right)\right)\\ ={\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{c}}_1\cos \left(4\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(4\mathrm{t}\right)\right)\end{array}$

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

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

 

  $\begin{array}{l} \mathrm{y}\left(0\right)={\mathrm{e}}^{-0}\left({\mathrm{c}}_1\cos (4\times 0)+{\mathrm{c}}_2\sin (4\times 0)\right)\\ {\mathrm{c}}_1=1\end{array}$

 

And

 $\mathrm{y}\text{'}\left(\mathrm{t}\right)=-{\mathrm{e}}^{-\mathrm{t}}({\mathrm{c}}_1\cos 4\mathrm{t}+{\mathrm{c}}_2\sin 4\mathrm{t})+{\mathrm{e}}^{-\mathrm{t}}(-4{\mathrm{c}}_1\mathrm{sint}+4{\mathrm{c}}_2\mathrm{cost})$

 

 

Then,

 $\begin{array}{l} \mathrm{y}\text{'}\left(0\right)=-{\mathrm{e}}^{-0}\left({\mathrm{c}}_1\cos (4\times 0)+{\mathrm{c}}_2\sin (4\times 0)\right)+{\mathrm{e}}^{-0}(-4{\mathrm{c}}_1\sin (4\times 0)+4{\mathrm{c}}_2\cos (4\times 0))\\ -{\mathrm{c}}_1+4{\mathrm{c}}_2=-1\end{array}$

 

Substitute $c_1$  in the above equation

 

 $\begin{array}{c}-1+{\mathrm{c}}_2=-1\\ {\mathrm{c}}_2=0\end{array}$

 

Therefore, the solution is  $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left(\cos 4\mathrm{t}\right)$.