Answer:
To solve the given second-order linear homogeneous differential equation:
y" + 2y' + 2y = u1(t) - u2(t)
where y(0) = 0 and y'(0) = 0, we can follow these steps:
Step 1: Solving the homogeneous equation
First, let's solve the homogeneous part of the equation, where u1(t) - u2(t) is set to zero:
y" + 2y' + 2y = 0
The characteristic equation for this homogeneous equation is given by:
r^2 + 2r + 2 = 0
Using the quadratic formula, we find the roots of the characteristic equation:
r = (-2 ± √(2^2 - 4*1*2)) / 2
r = (-2 ± √(-4)) / 2
r = -1 ± i√3
So, the general solution to the homogeneous equation is:
y_h(t) = c1 * e^(-t) * cos(√3t) + c2 * e^(-t) * sin(√3t)
Step 2: Finding the particular solution
Now, we need to find a particular solution for the non-homogeneous part of the equation, which is u1(t) - u2(t). This is a step function with a change at t = 0.
For t < 0, both u1(t) and u2(t) are zero, so the equation becomes:
y" + 2y' + 2y = 0
Using the initial conditions y(0) = 0 and y'(0) = 0, we can solve for the constants c1 and c2.
At t = 0, we have:
y_h(0) = c1 * e^0 * cos(0) + c2 * e^0 * sin(0) = c1 = 0
Taking the derivative at t = 0:
y_h'(t) = -c1 * e^(-t) * cos(√3t) * √3 + c2 * e^(-t) * sin(√3t) * √3
y_h'(0) = -c1 * √3 + c2 * 0 = 0
=> c1 = 0
So, for the homogeneous part, we have y_h(t) = 0.
Step 3: Considering the non-homogeneous part
Now, let's consider the non-homogeneous part u1(t) - u2(t) for t ≥ 0.
At t ≥ 0, u1(t) = 1 and u2(t) = 0, so the equation becomes:
y" + 2y' + 2y = 1
To find the particular solution for this equation, we can assume a constant solution, y_p(t) = A, where A is a constant.
Plugging this into the equation, we get:
0 + 0 + 2A = 1
2A = 1
A = 1/2
So, the particular solution is y_p(t) = 1/2.
Step 4: Combining the general and particular solutions
The general solution for the entire equation is given by the sum of the homogeneous and particular solutions:
y(t) = y_h(t) + y_p(t)
= 0 + 1/2
= 1/2
Therefore, the solution to the differential equation y" + 2y' + 2y = u1(t) - u2(t), with y(0) = 0 and y'(0) = 0, is:
y(t) = 1/2