Solve:
x'=x+y+2t
y'=x+y-2t
To solve this system of differential equations, you can use the method of finding the eigenvalues and eigenvectors of the coefficient matrix. Let's start by rewriting the system in matrix form:
X' = AX + Bt
where X = [x, y], A = [[1, 1], [1, 1]], and B = [2, -2].
To find the eigenvalues, we solve the characteristic equation:
|A - λI| = 0
where I is the identity matrix. Substituting the values into the characteristic equation:
|[[1, 1], [1, 1]] - λ[[1, 0], [0, 1]]| = 0
We get:
|(1-λ, 1), (1, 1-λ)| = 0
Expanding the determinant:
(1-λ)(1-λ) - 1 = 0
(λ - 2)(λ + 1) = 0
So, we have two eigenvalues: λ1 = 2 and λ2 = -1.
Next, we find the eigenvectors corresponding to each eigenvalue.
For λ1 = 2, we solve the equation (A - λ1I)v1 = 0:
[[1-2, 1], [1, 1-2]]v1 = 0
[[-1, 1], [1, -1]]v1 = 0
We obtain the system of equations:
-v1 + v2 = 0
v1 - v2 = 0
The general solution is v1 = t[1, 1], where t is a scalar parameter.
For λ2 = -1, we solve the equation (A - λ2I)v2 = 0:
[[1+1, 1], [1, 1+1]]v2 = 0
[[2, 1], [1, 2]]v2 = 0
This system of equations simplifies to:
2v1 + v2 = 0
v1 + 2v2 = 0
The general solution is v2 = s[-1, 2], where s is a scalar parameter.
Now, we can write the general solution to the system of differential equations:
X = c1e^(λ1t)v1 + c2e^(λ2t)v2
where c1 and c2 are constants.
Substituting the values for λ1, λ2, v1, and v2:
X = c1e^(2t)[1, 1] + c2e^(-t)[-1, 2]
X = [c1e^(2t) - c2e^(-t), c1e^(2t) + 2c2e^(-t)]
This is the general solution to the system of differential equations.
To find the particular solution, you need initial conditions for x and y at some given time t0.