Question

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

Answer 1

The solution of the problem is
`y(t) = e^(2 t) cos(2 t)`

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 = (4+√(-16) )/(2)` or
`x = (4-√(-16) )/(2)`

`x = (2+4i )/(2)` or
`x = (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)`