Answer:
y(t) = 3/4exp(4t) - 1/4exp(-4t) - 1/2
z(t) = -3/4exp(4t) - 1/4exp(-4t) - 2
Explanation:
Given system of differential equations:
dy/dt = 2y + 2z
dz/dt = 2y + 2z
We can write this system in matrix form as:
d/dt [y z] = [2 2] [y z]
Let A = [2 2]. Then the system can be written as:
d/dt [y z] = A[y z]
The solution to this system is given by:
[y z] = exp(At) [y(0) z(0)]
where exp(At) is the matrix exponential of At.
To find exp(At), we first need to find the eigenvalues and eigenvectors of A. The characteristic equation of A is:
det(A - lambdaI) = 0
=> det([2-lambda 2; 2 2-lambda]) = 0
=> (2-lambda)(2-lambda) - 4 = 0
=> lambda1 = 4, lambda2 = 0
The eigenvectors corresponding to lambda1 = 4 and lambda2 = 0 are:
v1 = [1 1] and v2 = [-1 1]
We can now write A as:
A = PDP^-1
where P = [v1 v2] and D = [4 0; 0 0]. Then,
exp(At) = Pexp(Dt)P^-1
We can compute exp(Dt) as:
exp(Dt) = [exp(4t) 0; 0 1]
Therefore,
exp(At) = [1/2 1/2; -1/2 1/2] [exp(4t) 0; 0 1] [1/2 -1/2; 1/2 1/2]
Now, we can find the solution to the system as:
[y z] = exp(At) [y(0) z(0)]
=> [y z] = [1/2 1/2; -1/2 1/2] [exp(4t) 0; 0 1] [1/2 -1/2; 1/2 1/2] [-1; -2]
=> [y z] = [1/2 1/2; -1/2 1/2] [exp(4t) 0; 0 1] [3/2; -1/2]
=> [y z] = [1/2 1/2; -1/2 1/2] [3/2exp(4t); -1/2]
=> [y z] = [3/2exp(4t)/2 - 1/2exp(-4t)/2; -3/2exp(4t)/2 - 1/2exp(-4t)/2]
Therefore, the solution to the system of differential equations with the given initial conditions is:
y(t) = 3/4exp(4t) - 1/4exp(-4t) - 1/2
z(t) = -3/4exp(4t) - 1/4exp(-4t) - 2