Respuesta :
Answer:
The solution is [tex]y(t) = 3e^{-2 t}cos(6 t) + \frac{11}{6} e^{-2 t}sin(6 t)[/tex]
Step-by-step explanation:
First we will write the characteristic equation, which is
[tex]x^{2} + 4x + 40 = 0[/tex]
This is a quadratic equation
For the general form of quadratic equation, [tex]ax^{2} +bx + c = 0[/tex], [tex]x[/tex] is given by the general formula,
[tex]x = \frac{-b+\sqrt{b^{2} - 4ac } }{2a}[/tex] or [tex]x = \frac{-b-\sqrt{b^{2} - 4ac } }{2a}[/tex]
Hence, for [tex]x^{2} + 4x + 40 = 0[/tex]
[tex]a = 1, b = 4,[/tex] and [tex]c = 40[/tex]
Hence, the formula becomes
[tex]x = \frac{-4+\sqrt{(4)^{2} - 4(1)(40) } }{2(1)}[/tex] or [tex]x = \frac{-4-\sqrt{(4)^{2} - 4(1)(40) } }{2(1)}[/tex]
[tex]x = \frac{-4+\sqrt{16 - 160} }{2}[/tex] or [tex]x = \frac{-4-\sqrt{16 - 160} }{2}[/tex]
[tex]x = \frac{-4+\sqrt{-144} }{2}[/tex] or [tex]x = \frac{-4-\sqrt{-144} }{2}[/tex]
[tex]x = \frac{-4+ 12i }{2}[/tex] or [tex]x = \frac{-4- 12i }{2}[/tex]
[tex]x = -2 + 6i[/tex] or [tex]x = -2 - 6i[/tex]
Hence, [tex]x = -2[/tex] ± [tex]6i[/tex]
That is, [tex]x_{1} = -2 + 6i[/tex] or [tex]x_{2} = -2 - 6i[/tex]
These are the roots of the equation.
For the general solution,
Since the roots of the equation are complex,
If the roots of a characteristic equation are in the form [tex](\alpha[/tex] ± [tex]\beta i)[/tex], then the general solution is given by
[tex]y(t) = C_{1}e^{\alpha t}cos(\beta t) + C_{2}e^{\alpha t}sin(\beta t)[/tex]
Hence, the general solution for the differential equation becomes
[tex]y(t) = C_{1}e^{-2 t}cos(6 t) + C_{2}e^{-2 t}sin(6 t)[/tex]
Then
[tex]y'(t) = -2C_{1}e^{-2t} cos(6t) - 6C_{1}e^{-2t} sin(6t) -2C_{2}e^{-2t}sin(6t) + 6C_{2}e^{-2t} cos(6t)[/tex]
Now, from the question,
y(0)= 3
y′(0)=5.
That is,
[tex]3 = y(0) = C_{1}e^{-2(0)}cos(\beta (0)) + C_{2}e^{-2(0)}sin(\beta (0))[/tex]
[tex]3 = C_{1}e^{0}cos(0) + C_{2}e^{0}sin(0)[/tex]
[tex]3 = C_{1}[/tex]
∴[tex]C_{1} = 3[/tex]
[NOTE: [tex]e^{0} = 1[/tex] and [tex]cos(0) = 1[/tex] and [tex]sin(0) = 0[/tex] ]
Also,
[tex]5 = y'(0) = -2C_{1}e^{-2(0)} cos(6(0)) - 6C_{1}e^{-2(0)} sin(6(0)) -2C_{2}e^{-2(0)}sin(6(0)) + 6C_{2}e^{-2(0)} cos(6(0))[/tex]
[tex]5 = -2C_{1} + 6C_{2}[/tex]
Now, we can find [tex]C_{2}[/tex] by putting the value of [tex]C_{1} = 3[/tex] into the equation
[tex]5 = -2C_{1} + 6C_{2}[/tex]
[tex]5 = -2(3) + 6C_{2}[/tex]
[tex]5 = -6 + 6C_{2}\\11 = 6C_{2}\\C_{2} = \frac{11}{6}[/tex]
∴ [tex]C_{1} = 3[/tex] and [tex]C_{2} = \frac{11}{6}[/tex]
Then the solution becomes
[tex]y(t) = 3e^{-2 t}cos(6 t) + \frac{11}{6} e^{-2 t}sin(6 t)[/tex]
