y''-4y'+3y=0 y(0)=0 and y'(0)=-1
We can solve this by letting the general solution be
y(t) = C1 e r1t + C2 e r2t
And we can find r1 and r2 thusly
(r - 3) ( r - 1) = 0 setting both factors to 0 and solving for r we have that r1 = 3 and r2 = 1
So......the general solution is :
y(t) = C1 e3t + C2 e1t and
y' (t) = 3Ci e 3t + C2 e 1t
Applying the initial conditions, we have that
0 = C1 + C2 → C2 = - C1 (1)
-1 = 3C1 + C2 (2)
Subbing (1) into (2) we have that
-1 = 3C1 - C1
-1 = 2C1
-1/2 = C1 → C2 = -C1 = 1/2
So......the actual solution is
y(t) = (-1/2) e 3t + (1/2) e1t