Solve the system of differential equations: dot x=-2x+y-2z…
Solve the system of differential equations: $$ \dot x=-2x+y-2z,\qquad \dot y=x-2y+2z,\qquad \dot z=3x-3y+5z $$
Explanation of Method
The characteristic cubic has a repeated root $\lambda=-1$; since $A+I$ has rank 1 (a full two-dimensional null space), no generalized eigenvector is needed and the matrix is completely diagonalisable, alongside one simple eigenvalue $\lambda=3$.
- Write the system in matrix form and compute the characteristic polynomial.
- Solve the cubic for the eigenvalues.
- Find the eigenvector for the simple eigenvalue.
- Check the geometric multiplicity of the repeated eigenvalue and find its eigenspace.
- Assemble and verify the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}-2&1&-2\\1&-2&2\\3&-3&5\end{pmatrix}$, expanding $\det(A-\lambda I)$ gives
$$ -\lambda^3+\lambda^2+5\lambda+3=-(\lambda+1)^2(\lambda-3)=0. $$
Eigenvalues: $\lambda_1=-1$ (algebraic multiplicity 2) and $\lambda_2=3$ (simple).
Step 2 · Eigenvector for $\lambda_2=3$
$$ \mathbf v_3=\begin{pmatrix}-1\\1\\3\end{pmatrix},\qquad A\mathbf v_3=3\mathbf v_3.\ \checkmark $$
Step 3 · Eigenspace for $\lambda_1=-1$
$A+I=\begin{pmatrix}-1&1&-2\\1&-1&2\\3&-3&6\end{pmatrix}$ has rank 1 (rows 2 and 3 are scalar multiples of row 1), so the single equation $v_1=v_2-2v_3$ leaves two free parameters — geometric multiplicity 2, equal to the algebraic multiplicity. Two independent eigenvectors:
$$ \mathbf p=\begin{pmatrix}1\\1\\0\end{pmatrix},\qquad \mathbf q=\begin{pmatrix}-2\\0\\1\end{pmatrix}, $$
each verified directly: $A\mathbf p=-\mathbf p$, $A\mathbf q=-\mathbf q$. Since geometric multiplicity equals algebraic multiplicity, $A$ is diagonalisable at $\lambda=-1$ — no Jordan block is needed.
Step 4 · Linear independence and general solution
The determinant of $[\mathbf p\ \mathbf q\ \mathbf v_3]$ is $4\neq0$, so the three eigenvectors are independent and superpose directly.
Step 5 · Verification
Substituting the general solution into all three original equations and matching the coefficients of $e^{-t}$ and $e^{3t}$ confirms every equation holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda=-1$ (mult.\ 2), $\lambda=3$ (simple) |
| Geometric multiplicity at $\lambda=-1$ | 2 (equals algebraic multiplicity) |
| Structure | diagonalisable, no Jordan blocks |
Additional Notes
- The full-rank-1 structure of $A+I$ (all three rows proportional) is what guarantees the two-dimensional eigenspace and rules out a Jordan block, without needing to hunt for a generalized eigenvector.
- As $t\to\infty$ the $e^{3t}$ term dominates unless $C_3=0$, in which case the solution decays to zero along the $\lambda=-1$ eigenplane.
- Verified by direct substitution into all three original equations.
Result
The general solution of the system is: