Second order differential equations?

Question

I'm having problem with some basic 2nd order equations. The problem states, find the solution of the given initial value problem: y'' + 3y' = 0; y(0) = -2 and y'(0) = 3. The ANSWER is y = -1 = e^(-3t). Thanks for the help.

Answer 1

Easiest way to attack this type of problem is integrate once. So

y"+3y' = 0 --> integrate y'+3y = c =constant

Now trial solution y = Ae^(-3t) so
(-3A+3)e^(3-t)=c

Need to find A and c. Apply initial conditions:

y(0) = -2 = -3A+A-c
y'(0) = (9A-9)=0 ---> A=1

So c =0

Thus y(t) = e^(-3t)