1. **Stating the problem:** We are given a system of three linear differential equations: $$\begin{cases} x' = 2x + y + 2z \\ y' = x + 2y + 2z \\ z' = x + y + 3z \end{cases}$$ We want to analyze and solve this system. 2. **Matrix form:** Rewrite the system in matrix form as: $$\mathbf{x}' = A\mathbf{x}$$ where $$\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{bmatrix}$$ 3. **Finding eigenvalues:** To solve, find eigenvalues $\lambda$ by solving: $$\det(A - \lambda I) = 0$$ where $I$ is the identity matrix. Calculate: $$\det\begin{bmatrix} 2-\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix} = 0$$ Expanding the determinant: $$ (2-\lambda) \left((2-\lambda)(3-\lambda) - 2 \cdot 1\right) - 1 \left(1(3-\lambda) - 2 \cdot 1\right) + 2 \left(1 \cdot 1 - (2-\lambda) \cdot 1\right) = 0 $$ Simplify step-by-step: - Compute inner terms: $$ (2-\lambda)(3-\lambda) = 6 - 2\lambda - 3\lambda + \lambda^2 = 6 - 5\lambda + \lambda^2 $$ - So, $$ (2-\lambda)(3-\lambda) - 2 = (6 - 5\lambda + \lambda^2) - 2 = 4 - 5\lambda + \lambda^2 $$ - Next terms: $$ 1(3-\lambda) - 2 = 3 - \lambda - 2 = 1 - \lambda $$ $$ 1 \cdot 1 - (2-\lambda) \cdot 1 = 1 - (2 - \lambda) = 1 - 2 + \lambda = \lambda - 1 $$ Putting all together: $$ (2-\lambda)(4 - 5\lambda + \lambda^2) - 1(1 - \lambda) + 2(\lambda - 1) = 0 $$ Expand: $$ (2-\lambda)(\lambda^2 - 5\lambda + 4) - (1 - \lambda) + 2(\lambda - 1) = 0 $$ Multiply: $$ 2(\lambda^2 - 5\lambda + 4) - \lambda(\lambda^2 - 5\lambda + 4) - 1 + \lambda + 2\lambda - 2 = 0 $$ Simplify: $$ 2\lambda^2 - 10\lambda + 8 - (\lambda^3 - 5\lambda^2 + 4\lambda) - 1 + \lambda + 2\lambda - 2 = 0 $$ Distribute minus: $$ 2\lambda^2 - 10\lambda + 8 - \lambda^3 + 5\lambda^2 - 4\lambda - 1 + \lambda + 2\lambda - 2 = 0 $$ Combine like terms: - Cubic: $-\lambda^3$ - Quadratic: $2\lambda^2 + 5\lambda^2 = 7\lambda^2$ - Linear: $-10\lambda - 4\lambda + \lambda + 2\lambda = -11\lambda$ - Constants: $8 - 1 - 2 = 5$ So: $$ -\lambda^3 + 7\lambda^2 - 11\lambda + 5 = 0 $$ Multiply both sides by $-1$: $$ \lambda^3 - 7\lambda^2 + 11\lambda - 5 = 0 $$ 4. **Solving the cubic:** Try rational roots using factors of 5: $\pm1, \pm5$ Test $\lambda=1$: $$1 - 7 + 11 - 5 = 0$$ So $\lambda=1$ is a root. Divide polynomial by $(\lambda - 1)$: $$ \lambda^3 - 7\lambda^2 + 11\lambda - 5 = (\lambda - 1)(\lambda^2 - 6\lambda + 5) $$ Solve quadratic: $$ \lambda^2 - 6\lambda + 5 = 0 $$ Using quadratic formula: $$ \lambda = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2} $$ So: $$ \lambda = 5 \quad \text{or} \quad \lambda = 1 $$ Eigenvalues are: $$ \lambda_1 = 1, \quad \lambda_2 = 1, \quad \lambda_3 = 5 $$ 5. **Interpretation:** The system has eigenvalues 1 (with multiplicity 2) and 5. 6. **General solution form:** The solution is a linear combination of eigenvectors multiplied by $e^{\lambda t}$: $$ \mathbf{x}(t) = c_1 e^{t} \mathbf{v}_1 + c_2 t e^{t} \mathbf{v}_2 + c_3 e^{5t} \mathbf{v}_3 $$ where $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are eigenvectors corresponding to eigenvalues. 7. **Summary:** We transformed the system into matrix form, found eigenvalues by solving the characteristic polynomial, and identified the general solution structure involving exponentials of eigenvalues times eigenvectors. This completes the solution for the given system.