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 differential equation and then apply the initial conditions.
Step 1: Finding the general solution
The given differential equation is a second-order linear homogeneous equation. To find the general solution, we assume the solution has the form: x(t) = e^(rt). Here, r is a constant to be determined.
Substituting this into the differential equation, we have:
d^2x/dt^2 + 2dx/dt - 3x = 0
Differentiating x(t) twice and substituting back into the equation, we get:
r^2e^(rt) + 2re^(rt) - 3e^(rt) = 0
Factoring out e^(rt), we have:
e^(rt)(r^2 + 2r - 3) = 0
Since e^(rt) does not equal zero for any value of t, we are left with the quadratic equation:
r^2 + 2r - 3 = 0
Solving this equation, we find r = 1 and r = -3.
Thus, the general solution to the differential equation is:
x(t) = C₁e^(t) + C₂e^(-3t), where C₁ and C₂ are constants.
Step 2: Applying the initial conditions
To find the specific solution, we apply the initial conditions x(2π) = 1 and x'(2π) = 13.
x(2π) = 1:
Plugging in t = 2π into the general solution, we get:
1 = C₁e^(2π) + C₂e^(-6π)
x'(t) = dx(t)/dt
Taking the derivative of the general solution, we have:
x'(t) = C₁e^(t) - 3C₂e^(-3t)
x'(2π) = 13:
Plugging in t = 2π into the derivative of the general solution, we get:
13 = C₁e^(2π) - 3C₂e^(-6π)
Now we have a system of two equations:
1 = C₁e^(2π) + C₂e^(-6π)
13 = C₁e^(2π) - 3C₂e^(-6π)
We can solve this system of equations to find the values of C₁ and C₂.
Once we have the values of C₁ and C₂, we can substitute them back into the general solution to obtain the specific solution to the initial value problem.