Question

Solve the initial value problem

y′′+4y′+40y=0

y(0)= 3
y′(0)=5.

Answer 1

The solution is
`y(t) = 3e^(-2 t)cos(6 t) + (11)/(6) e^(-2 t)sin(6 t)`

Explanation:

First we will write the characteristic equation, which is

`x^(2) + 4x + 40 = 0`

This is a quadratic equation

For the general form of quadratic equation,
`ax^(2) +bx + c = 0`,
`x` is given by the general formula,

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

Hence, for
`x^(2) + 4x + 40 = 0`

`a = 1, b = 4,` and
`c = 40`

Hence, the formula becomes

`x = \frac{-4+\sqrt{(4)^(2) - 4(1)(40) } }{2(1)}` or
`x = \frac{-4-\sqrt{(4)^(2) - 4(1)(40) } }{2(1)}`

`x = (-4+√(16 - 160) )/(2)` or
`x = (-4-√(16 - 160) )/(2)`

`x = (-4+√(-144) )/(2)` or
`x = (-4-√(-144) )/(2)`

`x = (-4+ 12i )/(2)` or
`x = (-4- 12i )/(2)`

`x = -2 + 6i` or
`x = -2 - 6i`

Hence,
`x = -2` ±
`6i`

That is,
`x_(1) = -2 + 6i` or
`x_(2) = -2 - 6i`

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
`(\alpha` ±
`\beta i)`, then the general solution is given by

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

Hence, the general solution for the differential equation becomes

`y(t) = C_(1)e^(-2 t)cos(6 t) + C_(2)e^(-2 t)sin(6 t)`

Then

`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)`

Now, from the question,

y(0)= 3

y′(0)=5.

That is,

`3 = y(0) = C_(1)e^(-2(0))cos(\beta (0)) + C_(2)e^(-2(0))sin(\beta (0))`

`3 = C_(1)e^(0)cos(0) + C_(2)e^(0)sin(0)`

`3 = C_(1)`

∴
`C_(1) = 3`

[NOTE:
`e^(0) = 1` and
`cos(0) = 1` and
`sin(0) = 0` ]

Also,

`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))`

`5 = -2C_(1) + 6C_(2)`

Now, we can find
`C_(2)` by putting the value of
`C_(1) = 3` into the equation

`5 = -2C_(1) + 6C_(2)`

`5 = -2(3) + 6C_(2)`

`5 = -6 + 6C_(2)\\11 = 6C_(2)\\C_(2) = (11)/(6)`

∴
`C_(1) = 3` and
`C_(2) = (11)/(6)`

Then the solution becomes

`y(t) = 3e^(-2 t)cos(6 t) + (11)/(6) e^(-2 t)sin(6 t)`