Find the general solution of the following system of equation?

Please help, i don't know what to do...
dx/dt = 3x +2y +2z
dy/dt = -5x - 4y - 2z
dz/dt = 5x + 5y +3z.

I assume you covered the eigenvalues method in school for solution of system of differential equations:

Given:
x'=3x+2y+2z
y'=-5x-4y-2z
z'=5x+5y+3z

Using the eigenvalue method, and using uppercase symbols to represent matrices and/vectors, we have

X'=AX
[ 3 2 2 ]
[-5 -4 -2] X
[ 5 5 3 ]
X'=[x',y'z'] and
X=[x,y,z]

From which the eigenvalues of A can be found to be
λ=[3,-2,1]
where λ1=3, λ2=-2, λ3=1.

There will be three independent solutions to the homogeneous equation, each associated with an arbitrary integration constant named C1, C2 and C3 respectively.

Each solution will be associated with a vector K whose elements apply to each of the dependent variable x,y and z, such that:

X=CKe^(λt)
where
C=[C1,C2,C3], arbitrary constants
and
K=[k1,k2,k3] solution vector for x,y & z respectively.

For λ=3,
substitute λ=3 in the matrix A to get A-λ1I
[ 3-3 2 2 ]
[-5 -4-3 -2]
[ 5 5 3-3]
from which we get
[ 0 2 2 ]
[-5 -7 -2] K1=0
[ 5 5 0]
From the above matrix, we solve for k1 and k3 in terms of k2 to get
K1=[-1 1 -1]

Similarly, for λ=-2, we get
K2=[0,-1,1]
and
for λ=1, we get
K3=[1,-1,0]

Therefore the final solution of
X=CKe^(λt) is

[x,y,z]=
C1*K1*e^(3t) + C2*K2*e^(-2t) + C3*K3*e^t

Separating the equation into three scalar equations, we get by extracting the individual values from K1, K2 and K3,

x(t)=-C1*e^(3t)+C3*e^t
y(t)=2C1*e^(3t)-C2*e^(-2t)-C3*e^t
z(t)=-C1*e^(3t)+C2*e^(-2t)

I suggest you substitute these solutions into the original equations to make sure they equal to the left hand side, x'(t), y'(t), etc.

To find the general solution of this system of equations, we can use matrix methods. Let's express the given system of equations in matrix form.

The system of equations can be represented as:

\(\begin{bmatrix} \frac{{dx}}{{dt}} \\ \frac{{dy}}{{dt}} \\ \frac{{dz}}{{dt}} \end{bmatrix} = \begin{bmatrix} 3 & 2 & 2 \\ -5 & -4 & -2 \\ 5 & 5 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}\)

In this matrix equation, the coefficient matrix on the right side is denoted as \(A\) and the column matrix of variables is denoted as \(X\). So we have:

\(\frac{{dX}}{{dt}} = AX\)

To find the general solution, we can solve the associated homogeneous linear system by finding the eigenvalues and eigenvectors of the matrix \(A\).

Step 1: Find the eigenvalues of matrix \(A\):
To find the eigenvalues, we solve the characteristic equation: det(\(A - \lambda I\)) = 0.
Substituting the values from matrix \(A\) into the characteristic equation, we get:

\(\begin{vmatrix} 3 - \lambda & 2 & 2 \\ -5 & -4 - \lambda & -2 \\ 5 & 5 & 3 - \lambda \end{vmatrix} =0\)

Expanding the determinant and simplifying, we have:

\((3 - \lambda)((-4 - \lambda)(3 - \lambda) + 10) - 2((-5)(3 - \lambda) - (5)(-2)) = 0\)

Simplifying further, we have:

\(-(4 + \lambda)(\lambda^2 - 2\lambda - 4) + 10(3 - \lambda) - 2(-15 + 10) = 0\)

\((4 + \lambda)(\lambda^2 - 2\lambda - 4) - 20\lambda + 50 = 0\)

Expanding and arranging the terms, we get:

\(\lambda^3 - 2\lambda^2 - 4\lambda + 4\lambda^2 - 8\lambda - 16 -20\lambda + 50 = 0\)

Combining like terms, we have:

\(\lambda^3 + 2\lambda^2 - 36\lambda + 34 = 0\)

Solving this equation may require numerical methods or factoring.

Step 2: Find the eigenvectors associated with each eigenvalue.
Once we find the eigenvalues, we can find the eigenvectors by solving the system of equations: \(AX = \lambda X\), where \(\lambda\) is one of the eigenvalues found.

For each eigenvalue, substitute it back into the matrix equation \(AX = \lambda X\) and solve the resulting system of equations. Each solution will correspond to an eigenvector.

Step 3: Write the general solution.
Once we have found the eigenvalues and eigenvectors, the general solution can be expressed as a linear combination of the eigenvectors.

So, the general solution to the given system of equations is:

\(X = c_1e^{\lambda_1 t}V_1 + c_2e^{\lambda_2 t}V_2 + c_3e^{\lambda_3 t}V_3\)

where \(c_1, c_2, c_3\) are constants, \(\lambda_1, \lambda_2, \lambda_3\) are the eigenvalues, and \(V_1, V_2, V_3\) are the corresponding eigenvectors.