Solve the initial value problem
d^2x/dt^2+2dx/dt-3x = 0, x(2π) =1, x(2π) = 13
To solve the given initial value problem, we need to find the solution to the second-order linear homogeneous differential equation:
d^2x/dt^2 + 2(dx/dt) - 3x = 0
First, let's find the characteristic equation for this differential equation. Assume a solution given by:
x(t) = e^(rt)
Now, differentiate x(t) twice with respect to t:
dx/dt = re^(rt)
d^2x/dt^2 = r^2e^(rt)
Substitute these derivatives into the differential equation:
r^2e^(rt) + 2(re^(rt)) - 3e^(rt) = 0
Factoring out e^(rt):
e^(rt)(r^2 + 2r - 3) = 0
Since e^(rt) cannot be zero (nontrivial solution), we focus on the quadratic factor:
r^2 + 2r - 3 = 0
Now, solve this quadratic equation for r using factoring or the quadratic formula:
(r + 3)(r - 1) = 0
This gives two possible values for r:
r1 = -3, r2 = 1
Since we have distinct real roots, the general solution for x(t) is a linear combination of these solutions:
x(t) = C1e^(-3t) + C2e^t
To find the particular solution that satisfies the initial conditions, we need to substitute the values of x(2π) = 1 and x'(2π) = 13 into the general solution.
At t = 2π:
x(2π) = C1e^(-6π) + C2e^(2π) = 1
At t = 2π:
dx/dt = -3C1e^(-6π) + C2e^(2π) = 13
Now, we have a system of two equations that we can solve to find the coefficients C1 and C2.
Solving the first equation for C1:
C1 = (1 - C2e^(2π)) / e^(-6π)
Substituting the value of C1 into the second equation:
-3(1 - C2e^(2π)) / e^(-6π) + C2e^(2π) = 13
Simplifying the equation:
-3 + 3C2e^(2π) + C2e^(8π) = 13e^(-6π)
Rearranging terms:
(3e^(2π) + e^(8π))C2 = 13e^(-6π) + 3
Finally, solving for C2:
C2 = (13e^(-6π) + 3) / (3e^(2π) + e^(8π))
With the values of C1 and C2, we can substitute them back into the general solution:
x(t) = C1e^(-3t) + C2e^t
Now we have the solution to the initial value problem.