The roots of the auxiliary equation are -1 + i and -1 - i.
To solve the given initial value problem (IVP) and graph the solution, we can follow these steps:
Step 1: Write the auxiliary equation
The given differential equation is y'' + 2y' + 2y = 0. To solve it, we can write the auxiliary equation by assuming y = e^(rt), where r is a constant:
r^2 + 2r + 2 = 0
Step 2: Solve the auxiliary equation
To solve the quadratic equation, we can use the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / 2a
In our case, a = 1, b = 2, and c = 2. Substituting these values into the quadratic formula, we get:
r = (-2 ± √(2^2 - 4(1)(2))) / (2(1))
r = (-2 ± √(-4)) / 2
r = (-2 ± 2i) / 2
r = -1 ± i
So, the roots of the auxiliary equation are -1 + i and -1 - i.
Step 3: Write the general solution
Since the roots of the auxiliary equation are complex, the general solution can be written as:
y = C1e^((-1 + i)t) + C2e^((-1 - i)t)
Step 4: Apply initial conditions
We are given the initial conditions y(0) = 0 and y'(0) = 1. Let's use these conditions to find the values of C1 and C2.
Substituting t = 0, y = 0 into the general solution:
0 = C1e^0 + C2e^0
0 = C1 + C2
Next, differentiate the general solution with respect to t:
y' = -C1e^((-1 + i)t) - C2e^((-1 - i)t)
Substituting t = 0, y' = 1 into the derived equation:
1 = -C1e^0 - C2e^0
1 = -C1 - C2
Now, we have a system of equations:
0 = C1 + C2
1 = -C1 - C2
Solving this system, we find C1 = -1/2 and C2 = 1/2.
Step 5: Write the particular solution
Using the values of C1 and C2, we can write the particular solution as:
y = -1/2e^((-1 + i)t) + 1/2e^((-1 - i)t)
Step 6: Simplify the particular solution
To simplify the particular solution, we can use Euler's formula, which states that e^(it) = cos(t) + isin(t).
Using Euler's formula, we can rewrite the particular solution as:
y = -1/2e^(-t)cos(t) + i/2e^(-t)sin(t) + 1/2e^(-t)cos(t) - i/2e^(-t)sin(t)
Simplifying further, we get:
y = e^(-t)cos(t)
Step 7: Graph the solution
To graph the solution, we can plot the function y = e^(-t)cos(t) on a coordinate system.
The distance between successive roots of the solution can be computed by finding the values of t where y = 0. In this case, the roots occur at regular intervals since the function is not periodic. By calculating the difference between consecutive root values, we can verify if it is constant.