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**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.