Question:
Find the solution of the given initial value problem.
$$y"+4y'+5y = 0, \ y(0)=1, \ y'(0) = 0$$
Solution:
The characteristic equation of the given differential equation is $r^2 + 4r+5 = 0$.
$$r = \frac{-4 \pm \sqrt{16-20}}{2}$$
$$r = -2 \pm i$$
Therefore, the general solution is $y = c_1e^{-2t}cos \ t+c_2e^{-2t}sin \ t$
$$y'(t) = -c_1e^{-2t}(2cos\ t+sin \ t)+c_2e^{-2t}(cos \ t-2sin \ t)$$
Substitute the initial values in:
$$y(0) = c_1e^{0}cos \ 0+c_2e^{-0}sin \ 0 = 1$$
We get $c_1 = 1$.
$$y'(0) = -c_1e^{0}(2cos\ 0+sin \ 0)+c_2e^{0}(cos \ 0-2sin \ 0) = 0$$
$$ -2c_1+c_2 = 0$$
$$-2+c_2 = 0$$
$$c_2 = 2$$
Thus the general solution of the equation is $y = e^{-2t}cos \ t+2e^{-2t}sin \ t$
