Question

How to solve initial value problem?

y" + 2y' +y=0, y(0)=4 , y'(0)=-6

Answer 1

`y = 4e^-t - 2te^-t`

Explanation:

Our characteristic equation will be

`r^2 + 2r + 1 = 0`

`(r + 1)^2 = 0`

`r = -1`

Therefore our solution will be `y = c_1e^(-t) + c_2te^(-t)` (because there is only one real root). The last step is to solve for the values of `c_1` and `c_2`. We will need the derivative for this step.

`y' = -c_1e^(-t) + c_2(1)e^(-t) - c_2te^(-t)`

Our system will therefore be

`{(-6 = -c_1e^(0) + c_2e^0 - c_2(0)e^0), (4 = c_1e^0 + c_2(0)e^0):}`

Simplifying a bit, we get:

`{(-6 = c_2 - c_1), (c_1 = 4):}`

Thus our solution will be `c_1 = 4`, `c_2 = -2`.

The equation has solution `y = 4e^-t - 2te^-t`

Hopefully this helps!