Solve the system of differential equations: dot x=2x-y-z…
Solve the system of differential equations: $$ \dot x=2x-y-z,\qquad \dot y=3x-2y-3z,\qquad \dot z=-x+y+2z $$
Explanation of Method
One eigenvalue is zero (the matrix is singular), contributing a constant equilibrium term, while the repeated eigenvalue $\lambda=1$ again has full geometric multiplicity, so the whole system is diagonalisable with no secular terms.
- Write the system in matrix form and compute the characteristic polynomial.
- Identify $\lambda=0$ and the repeated eigenvalue $\lambda=1$.
- Find the eigenvector for $\lambda=0$ and check the eigenspace for $\lambda=1$.
- Assemble and verify the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}2&-1&-1\\3&-2&-3\\-1&1&2\end{pmatrix}$, expanding $\det(A-\lambda I)$ gives
$$ -\lambda(\lambda-1)^2=0. $$
Eigenvalues: $\lambda_1=0$ (simple) and $\lambda_2=1$ (algebraic multiplicity 2).
Step 2 · Eigenvector for $\lambda_1=0$
$$ \mathbf v_1=\begin{pmatrix}-1\\-3\\1\end{pmatrix},\qquad A\mathbf v_1=\mathbf 0.\ \checkmark $$
Step 3 · Eigenspace for $\lambda_2=1$
$A-I$ has rank 1 (all three rows are proportional to $(1,-1,-1)$), so $v_1=v_2+v_3$ leaves two free parameters: geometric multiplicity 2, equal to algebraic multiplicity. Two independent eigenvectors:
$$ \mathbf v_2=\begin{pmatrix}1\\1\\0\end{pmatrix},\qquad \mathbf v_3=\begin{pmatrix}1\\0\\1\end{pmatrix}. $$
$A$ is diagonalisable at $\lambda=1$ — no Jordan block.
Step 4 · General solution and verification
The three eigenvectors are independent ($\det=1\neq0$); substitution confirms the general solution below satisfies all three original equations. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=0$ (simple), $\lambda_2=1$ (mult.\ 2) |
| Null-space vector | $(-1,-3,1)$ |
| Structure | diagonalisable, no secular terms |
Additional Notes
- The $\lambda=0$ mode contributes a constant equilibrium vector $C_1(-1,-3,1)^\top$ lying in the null space of $A$.
- As $t\to\infty$ the solution grows like $e^t$ unless $C_2=C_3=0$.
- Verified by direct substitution into all three equations.
Result
The general solution of the system is: