![y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](/tpl/images/2659/4006/e1d76.png)
(See attached graph)
where the values of 
are the roots:
![y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]](/tpl/images/2659/4006/fb812.png)
where 
.![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]](/tpl/images/2659/4006/59a68.png)
.
, thus, ![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]](/tpl/images/2659/4006/0b5be.png)
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3](/tpl/images/2659/4006/ba1a7.png)
![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2](/tpl/images/2659/4006/a8727.png)
:
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](/tpl/images/2659/4006/e401b.png)
Section 5.2 Problem 21:
Solve the initial value problem and graph the solution.
<...
Mathematics, 16.02.2022 06:20 danejac99
Section 5.2 Problem 21:
Solve the initial value problem and graph the solution.
Answers: 1
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