Answer:
[tex]y(x) =e^{-2 x}(3cos6 x + \frac{11}{6} sin6 x) + K[/tex]
Step-by-step explanation:
Given the initial value problem y′′+4y′+40y=0 where y(0)= 3 and y′(0)=5.
Find the solution in the attached file.
Since the solution is a complex number, the solution to the second order differential equation is given as shown;
y(x) = [tex]e^{\alpha x}(c_1cos\beta x + c_2sin \beta x)[/tex]
from the complex solution D = -2+6i
[tex]\alpha = -2, \beta = 6[/tex]
y(x) = [tex]e^{-2 x}(c_1cos6 x + c_2sin6 x)[/tex]
Substituting the initial values
when y(0) = 3, the solution becomes;
[tex]3 = e^{-2 x}(c_1cos6 (0) + c_2sin 6 (0))\\[/tex]
[tex]3 = e^{-2 (0)}(c_1+ 0)\\3 = 1*c_1\\c_1 = 3[/tex]
From y(x) = [tex]e^{\alpha x}(c_1cos\beta x + c_2sin \beta x)[/tex]
[tex]y'(x) = e^{\alpha x}(-\beta c_1sin\beta x + \beta c_2cos \beta x)+\alpha e^{\alpha x}(c_1cos\beta x + c_2sin \beta x)[/tex]
at y'(0) = 5, the solution becomes;
[tex]y'(x) = e^{\alpha x}(-\beta c_1sin\beta x + \beta c_2cos \beta x)+\alpha e^{\alpha x}(c_1cos\beta x + c_2sin \beta x)\\5 = e^{-2 (0)}(-6 c_1sin6 (0) + 6 c_2cos 6 (0))+-2 e^{-2 (0)}(c_1cos6 (0) + c_2sin 6 (0))\\5 = 1*(0 + 6 c_2)-2(c_1 + 0)\\5 = 6c_2-2c_1[/tex]
Since c₁ = 3;
5 = 6c₂ - 2(3)
5 = 6c₂ - 6
6c₂ = 11
c₂ = 11/6
Substituting the constant into the solution to the differential equation [tex]y(x) =e^{-2 x}(c_1cos6 x + c_2sin6 x)[/tex]
[tex]y(x) =e^{-2 x}(3cos6 x + \frac{11}{6} sin6 x) + K[/tex]
