Find the solution of the given inticial value problem: y'' + 3y' = 0, y(0)=-2, y'(0) = 3.
For problems in form of $$ax^2+bx+c = 0$$, the solution is $$r =\frac{-b\pm{ \sqrt{b^2-4ac}}}{2a}$$
In this case: $$\begin{align*} r &=\frac{-3\pm{ \sqrt{3^2-4\times 1\times 0}}}{2} \\ &= \frac{-3\pm3}{2} \end{align*}$$
Therefore we have: $$ r = 0, r = -3 $$
The genersal solution is: $$ y=c_1+c_2e^{-3t}$$
Substitute t=0 into y: $$c_1+c_2 = -2$$
Substitute t=0 into y': $$-3c_2 = 3$$
Solve these two equation we have: $$c_1= -1, c_2 = -1$$
Finally: $$y=-e^{-3t}-1$$
