Solve the system of differential equations: dot x=4x-y-z…
Solve the system of differential equations: $$ \dot x=4x-y-z,\qquad \dot y=x+2y-z,\qquad \dot z=x-y+2z $$
Explanation of Method
The characteristic polynomial has a repeated root $\lambda=3$; because the corresponding eigenspace turns out to be two-dimensional (rank of $A-3I$ is 1), no generalized eigenvector is needed and the matrix is fully diagonalisable.
- Write the system in matrix form and compute the characteristic polynomial.
- Identify the repeated eigenvalue and the simple eigenvalue.
- Check the geometric multiplicity of the repeated eigenvalue.
- Assemble and verify the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}4&-1&-1\\1&2&-1\\1&-1&2\end{pmatrix}$, expanding $\det(A-\lambda I)$ gives
$$ -(\lambda-3)^2(\lambda-2)=0. $$
Eigenvalues: $\lambda=3$ (algebraic multiplicity 2) and $\lambda=2$ (simple).
Step 2 · Eigenvector for $\lambda=2$
$$ \mathbf v_1=\begin{pmatrix}1\\1\\1\end{pmatrix},\qquad A\mathbf v_1=2\mathbf v_1.\ \checkmark $$
Step 3 · Eigenspace for $\lambda=3$
$A-3I=\begin{pmatrix}1&-1&-1\\1&-1&-1\\1&-1&-1\end{pmatrix}$ has rank 1, so the single equation $v_1=v_2+v_3$ leaves two free parameters: the geometric multiplicity is 2, equal to the algebraic multiplicity. Two independent eigenvectors:
$$ \mathbf v_2=\begin{pmatrix}1\\1\\0\end{pmatrix},\qquad \mathbf v_3=\begin{pmatrix}1\\0\\1\end{pmatrix}. $$
Since geometric multiplicity equals algebraic multiplicity, $A$ is diagonalisable — no generalized eigenvector or secular ($te^{3t}$) term is needed.
Step 4 · General solution and verification
The three eigenvectors are linearly independent ($\det=-1\neq0$), and direct substitution confirms the general solution below satisfies all three original equations. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda=3$ (mult.\ 2), $\lambda=2$ (simple) |
| Geometric multiplicity at $\lambda=3$ | 2 (equals algebraic multiplicity) |
| Jordan blocks | none — fully diagonalisable |
Additional Notes
- The plane $v_1=v_2+v_3$ is entirely an eigenspace for $\lambda=3$; any two independent vectors in it work as a basis.
- As $t\to\infty$ the $e^{3t}$ terms dominate the $e^{2t}$ contribution.
- Verified by direct substitution into all three equations.
Result
The general solution of the system is: