Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz4 => Topic started by: Yue Sagawa on October 18, 2019, 02:37:11 PM

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 $r^2 + 4r+5 = 0$.
$$r = \frac{4 \pm \sqrt{1620}}{2}$$
$$r = 2 \pm i$$
Therefore, the general solution is $y = c_1e^{2t}cos \ t+c_2e^{2t}sin \ t$
$$y'(t) = c_1e^{2t}(2cos\ t+sin \ t)+c_2e^{2t}(cos \ t2sin \ t)$$
Substitute the initial values in:
$$y(0) = c_1e^{0}cos \ 0+c_2e^{0}sin \ 0 = 1$$
We get $c_1 = 1$.
$$y'(0) = c_1e^{0}(2cos\ 0+sin \ 0)+c_2e^{0}(cos \ 02sin \ 0) = 0$$
$$ 2c_1+c_2 = 0$$
$$2+c_2 = 0$$
$$c_2 = 2$$
Thus the general solution of the equation is $y = e^{2t}cos \ t+2e^{2t}sin \ t$