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Initial value problem in Differential Equations

The following problem was given as an example by the professor but I can't seem to come up with the same answer as he did.
The answer he got was C1 = -2 and C2 = 1/4
I keep getting C1 = - 1/4 and C2 = - 1/2
There are three possibilities 1) He's wrong 2) I'm wrong or 3) we are both wrong.
I need to know which it is.
Can you show me how you would solve the problem and give me the correct answer?

Problem:
X = C1 cos4t + C2 sin4t is a solution for the differential equation X" + 16X = 0

The initial conditions are: X(pi/2) = -2 and X'(pi/2) = 1

I have determined that:

X' = - C1 4sin4t + C2 4cos4t
X" = - C1 16cos4t - C2 16sin4t

and that the equation is a solution for the differential equation.

It is when I try to determine the Initial values that I disgree with the professors answer.

Thank you,
Allan Cook

Solution Preview

for x(pi/2) = -2, substituting pi/2 for t to x equation:
x(pi/2) = c1 cos(4*pi/2) + c2 sin(4*pi/2) = c1 cos(2pi) + c2 sin(2pi)
x(pi/2)= c1 * 1 + c2 * 0
x(pi/2) ...

Solution Summary

It investigates an initial value problem in Differential Equations

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