Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz3 => Topic started by: Xinqiao Li on October 11, 2019, 02:00:35 PM

Find the solution of the given initial problem:
$$y''+4y'+3y = 0, y(0) = 2, y'(0) = 1$$
Assume $y = e^{rt}$, then it must follow that r is the root of the characteristic polynomial
$$r^2+4r+3=0\\
(r+1)(r+3)=0$$
We have $r_1 = 1$ or $r_2 = 3$.
The general solution of the second order differential equation has the form of
$$y = c_1e^{r_1t} + c_2e^{r_2t}$$
Thus, we have
$$y = c_1e^{t} + c_2e^{3t}$$
The derivative of this general solution is
$$y' = c_1e^{t}  3c_2e^{3t}$$
To satisfy both initial conditions $y(0) = 2$ and $y'(0) = 1$,
We have $2 = c_1 + c_2$ and $1 = c_1 3c_2$
This gives us $c_1 = \frac{5}{2}$ and $c_2 = \frac{1}{2}$
Therefore, the solution of the initial value problem is
$$y = \frac{5}{2}e^{t} \frac{1}{2}e^{3t}$$