Respuesta :
Answer:
The solution of the problem is [tex]y(t) = e^{2 t} cos(2 t)[/tex]
Step-by-step explanation:
First we will write the characteristic equation which is
[tex]x^{2} -4x + 8 = 0[/tex]
Now, we will solve this quadratic equation using the general formula.
Given a quadratic equation of the form, [tex]ax^{2} +bx +c = 0[/tex], then
From the general formula,
[tex]x = \frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex] or [tex]x = \frac{-b-\sqrt{b^{2}-4ac } }{2a}[/tex]
From the characteristic equation, [tex]a = 1, b = -4,[/tex] and [tex]c = 8[/tex]
Hence,
[tex]x = \frac{-(-4)+\sqrt{(-4)^{2}-4(1)(8) } }{2(1)}[/tex] or [tex]x = \frac{-(-4)-\sqrt{(-4)^{2}-4(1)(8) } }{2(1)}[/tex]
[tex]x = \frac{4+\sqrt{-16} }{2}[/tex] or [tex]x = \frac{4-\sqrt{-16} }{2}[/tex]
[tex]x = \frac{2+4i }{2}[/tex] or [tex]x = \frac{2-4i }{2}[/tex]
[tex]x = 2 + 2i[/tex] or [tex]x = 2 - 2i[/tex]
That is, [tex]x[/tex] = [tex]2[/tex] ± [tex]2i[/tex]
Then, [tex]x_{1} = 2 + 2i[/tex] and [tex]x_{2} = 2 - 2i[/tex]
These are the roots of the characteristic equation
The roots of the characteristic equation are complex, that is, in the form
([tex]\alpha[/tex] ± [tex]\beta i[/tex]).
For the general solution,
If the roots of a characteristic equation are in the form ([tex]\alpha[/tex] ± [tex]\beta i[/tex]), 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]
From the characteristic equation,
[tex]\alpha = 2[/tex] and [tex]\beta = 2[/tex]
Then, the general solution becomes
[tex]y(t) = C_{1}e^{2 t} cos(2 t) + C_{2}e^{2 t} sin(2t)[/tex]
Now, we will determine [tex]y'(t)[/tex]
[tex]y'(t) = 2C_{1}e^{2 t} cos(2 t) - 2C_{1}e^{2t}sin(2t) + 2C_{2} e^{2t}sin(2t) +2C_{2}e^{2t}cos(2t)[/tex]
From the question,
y(0) = 1
and
y'(0) = 2
Then,
[tex]1 = y(0) = C_{1}e^{2 (0)} cos(2 (0)) + C_{2}e^{2 (0)} sin(2(0))[/tex]
[tex]1 = C_{1}e^{ 0} cos(0) + C_{2}e^{0} sin(0)[/tex]
(NOTE: [tex]e^{0} = 1, cos(0) = 1[/tex] and [tex]sin(0) = 0[/tex] )
Then,
[tex]1 = C_{1}[/tex]
∴[tex]C_{1} = 1[/tex]
Also,
[tex]2 = y'(0) = 2C_{1}e^{2 (0)} cos(2 (0)) - 2C_{1}e^{2(0)}sin(2(0)) + 2C_{2} e^{2(0)}sin(2(0)) +2C_{2}e^{2(0)}cos(2(0))[/tex][tex]2 = 2C_{1}e^{0} cos(0) - 2C_{1}e^{0}sin(0) + 2C_{2} e^{0}sin(0) +2C_{2}e^{0}cos(0)[/tex]
[tex]2 = 2C_{1} +2C_{2}[/tex]
Then,
[tex]1 = C_{1} +C_{2}[/tex]
[tex]C_{2} = 1 - C_{1}[/tex]
Recall, [tex]C_{1} = 1[/tex]
∴ [tex]C_{2} = 1 - 1 = 0[/tex]
[tex]C_{2} = 0[/tex]
Hence, the solution becomes
[tex]y(t) = e^{2 t} cos(2 t)[/tex]
