Question

Solve the initial-value problem: y" - 4y' + 8y = 0, y(0) = 1, y'(0) = 2.

Answer 1

Answer:

The solution of the problem is $y(t) = e^{2 t} cos(2 t)$

Step-by-step explanation:

First we will write the characteristic equation which is

$x^{2} -4x + 8 = 0$

Now, we will solve this quadratic equation using the general formula.

Given a quadratic equation of the form, $ax^{2} +bx +c = 0$, then

From the general formula,

$x = \frac{-b+\sqrt{b^{2}-4ac } }{2a}$ or $x = \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

From the characteristic equation, $a = 1, b = -4,$ and $c = 8$

Hence,

$x = \frac{-(-4)+\sqrt{(-4)^{2}-4(1)(8) } }{2(1)}$ or $x = \frac{-(-4)-\sqrt{(-4)^{2}-4(1)(8) } }{2(1)}$

$x = \frac{4+\sqrt{-16} }{2}$ or $x = \frac{4-\sqrt{-16} }{2}$

$x = \frac{2+4i }{2}$ or $x = \frac{2-4i }{2}$

$x = 2 + 2i$ or $x = 2 - 2i$

That is, $x$ =  $2$ ± $2i$

Then, $x_{1} = 2 + 2i$ and $x_{2} = 2 - 2i$

These are the roots of the characteristic equation

The roots of the characteristic equation are complex, that is, in the form

($\alpha$ ± $\beta i$).

For the general solution,

If the roots of a characteristic equation are in the form  ($\alpha$ ± $\beta i$), the general solution is given by

$y(t) = C_{1}e^{\alpha t} cos(\beta t) + C_{2}e^{\alpha t} sin(\beta t)$

From the characteristic equation,

$\alpha = 2$ and $\beta = 2$

Then, the general solution becomes

$y(t) = C_{1}e^{2 t} cos(2 t) + C_{2}e^{2 t} sin(2t)$

Now, we will determine $y'(t)$

$y'(t) = 2C_{1}e^{2 t} cos(2 t) - 2C_{1}e^{2t}sin(2t) + 2C_{2} e^{2t}sin(2t) +2C_{2}e^{2t}cos(2t)$

From the question,

y(0) = 1

and

y'(0) = 2

Then,

$1 = y(0) = C_{1}e^{2 (0)} cos(2 (0)) + C_{2}e^{2 (0)} sin(2(0))$

$1 = C_{1}e^{ 0} cos(0) + C_{2}e^{0} sin(0)$

(NOTE: $e^{0} = 1, cos(0) = 1$ and $sin(0) = 0$ )

Then,

$1 = C_{1}$

∴$C_{1} = 1$

Also,

$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))$$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)$

$2 = 2C_{1} +2C_{2}$

Then,

$1 = C_{1} +C_{2}$

$C_{2} = 1 - C_{1}$

Recall, $C_{1} = 1$

∴ $C_{2} = 1 - 1 = 0$

$C_{2} = 0$

Hence, the solution becomes

$y(t) = e^{2 t} cos(2 t)$