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SOLVE SECOND ORDER INITIAL VALUE PROBLEM. 

y'' +6y' +5y = 0 , y(0)=1 and y' (0) = 1

 Aug 19, 2019
 #1
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SOLVE SECOND ORDER INITIAL VALUE PROBLEM.
\(y'' +6y' +5y = 0 ,\ y(0)=1 \text{ and } y' (0) = 1\)

 

\(\text{Let $y=e^{rx}$} \\ \text{Let $y'=re^{rx}$} \\ \text{Let $y''=r^2e^{rx}$} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{y'' +6y' +5y} &=& \mathbf{0} \\\\ r^2e^{rx} +6re^{rx} +5e^{rx} &=& 0 \\ \underbrace{e^{rx}}_{\neq 0}\cdot \underbrace{ \left(r^2 +6r +5\right) }_{=0} &=& 0 \\ && \begin{array}{|rcll|} \hline r^2 +6r +5 &=& 0 \\ r &=& \dfrac{-6\pm \sqrt{36-4\cdot 5}}{2} \\ r &=& \dfrac{-6\pm \sqrt{16}}{2} \\ r &=& \dfrac{-6\pm 4}{2} \\\\ r_1 &=& \frac{-6+4}{2} \\ \mathbf{r_1} &=& \mathbf{-1} \\\\ r_2 &=& \frac{-6-4}{2} \\ \mathbf{r_2} &=& \mathbf{-5} \\ \\ \hline \end{array} \\\\ y &=& c_1\cdot e^{r_1\cdot x} + c_2\cdot e^{r_2\cdot x} \\ \mathbf{y} &=& \mathbf{c_1\cdot e^{-1\cdot x} + c_2\cdot e^{-5\cdot x} } & (1) \\ \mathbf{y'} &=& \mathbf{-c_1\cdot e^{-1\cdot x} -5\cdot c_2\cdot e^{-5\cdot x} } & (2) \\ \hline y &=& c_1\cdot e^{-x} + c_2\cdot e^{-5\cdot x} \quad | \quad y(0)=1,\ x=0,\ y=1\\ 1 &=& c_1\cdot e^{-0} + c_2\cdot e^{-5\cdot 0} \\ \mathbf{1} &=& \mathbf{c_1 + c_2} & (3) \\\\ y' &=& -c_1\cdot e^{-x} -5\cdot c_2\cdot e^{-5\cdot x} \quad | \quad y'(0)=1, \ x=0,\ y'=1\\ 1 &=& -c_1\cdot e^{-0} -5\cdot c_2\cdot e^{-5\cdot 0} \\ \mathbf{1} &=& \mathbf{-c_1 - 5c_2} & (4) \\\\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (3) & \mathbf{1} &=& \mathbf{c_1 + c_2} \\ (4) & \mathbf{1} &=& \mathbf{-c_1 - 5c_2} \\ \hline (3)+(4): & 1+1 &=& c_1 + c_2+-c_1 - 5c_2 \\ & 2 &=& c_2 - 5c_2 \\ &-4c_2 &=& 2 \quad | \quad :(-4) \\ & \mathbf{c_2} &=& \mathbf{-\dfrac{1}{2}} \\\\ & 1 &=& c_1 + c_2 \\ & 1 &=& c_1 -\dfrac{1}{2}\\ & c_1 &=& 1+\dfrac{1}{2} \\ & \mathbf{c_1} &=& \mathbf{ \dfrac{3}{2}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{ c_1\cdot e^{-1\cdot x} + c_2\cdot e^{-5\cdot x} } \\ \mathbf{y} &=& \mathbf{ \dfrac{3}{2}\cdot e^{-x} -\dfrac{1}{2}\cdot e^{-5\cdot x} } \\ \hline \end{array}\)

 

laugh

 Aug 19, 2019

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