Originally Posted by
latavee y = c1e4x + c2e-x is the general solution of the differential equation y'' - 3y' - 4y = 0 on (-infinity, infinity). Find a member of the family that is a solution with initial conditions y(0) = 1 and y'(0) = 2.
What is the value of c1:
What is the value of c2:
So...a=1, b=3and c=-4
using the quad equation.... 4+/-Sqrt16-16/2a
The discriminat equals 0, so will use -b/2a for m1 and m2
So where do I go from here to find c1 and c2? We briefly touched this in class and I was trying to figure it out myself.