# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: Xinqiao Li on October 11, 2019, 02:00:35 PM

Title: TUT0502 Quiz 3
Post 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
$$y = \frac{5}{2}e^{-t} -\frac{1}{2}e^{-3t}$$