Solve the system of differential equations: dot x=x-2y-z…
Solve the system of differential equations: $$ \dot x=x-2y-z,\qquad \dot y=-x+y+z,\qquad \dot z=x-z $$
Explanation of Method
Three distinct real eigenvalues — including one exactly zero — give a straightforward eigenvector decomposition with a genuine equilibrium direction alongside one decaying and one growing exponential mode.
- Write the matrix $A$ and factor the characteristic cubic.
- Find an eigenvector for each of the three distinct eigenvalues.
- Assemble the general solution.
- Verify and describe the long-term behaviour.
Step 1 · Characteristic equation
With $A=\begin{pmatrix}1&-2&-1\\-1&1&1\\1&0&-1\end{pmatrix}$, expanding gives
$$ -\lambda(\lambda+1)(\lambda-2)=0, $$
so $\lambda_1=0,\ \lambda_2=-1,\ \lambda_3=2$.
Step 2 · Eigenvectors
$$ \mathbf v_1=\begin{pmatrix}1\\0\\1\end{pmatrix}\ (\lambda_1=0),\qquad \mathbf v_2=\begin{pmatrix}0\\1\\-2\end{pmatrix}\ (\lambda_2=-1),\qquad \mathbf v_3=\begin{pmatrix}3\\-2\\1\end{pmatrix}\ (\lambda_3=2). $$
Step 3 · General solution
$$ x(t)=C_1+3C_3e^{2t},\qquad y(t)=C_2e^{-t}-2C_3e^{2t},\qquad z(t)=C_1-2C_2e^{-t}+C_3e^{2t}. $$
Step 4 · Verification
Substituting back into all three original equations confirms each holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=0,\ \lambda_2=-1,\ \lambda_3=2$ |
| Equilibrium direction | $(1,0,1)$ (null space of $A$) |
| Long-term behaviour | dominated by $e^{2t}$ unless $C_3=0$ |
Additional Notes
- The zero eigenvalue contributes the constant $C_1$ to both $x$ and $z$ (matching the null-space direction $(1,0,1)$), while $\lambda_2=-1$ and $\lambda_3=2$ give decaying and growing transients respectively.
- Unless the initial data happens to have $C_3=0$, the $e^{2t}$ mode eventually dominates all three components.
- Verified by direct substitution into all three original equations.
Result
The general solution of the system is: