Solve the system of differential equations: (dx)/(dt) = y +…
Solve the system of differential equations: $$ \frac{dx}{dt} = y + z,\qquad \frac{dy}{dt} = x + z,\qquad \frac{dz}{dt} = x + y $$
Explanation of Method
The $3\times3$ coefficient matrix has one simple eigenvalue and a repeated eigenvalue whose eigenspace is two-dimensional, so no generalised eigenvector is needed — three independent eigenvectors give the full solution directly.
- Write the matrix $A$ and solve $\det(A-\lambda I)=0$.
- Find eigenvectors for each eigenvalue (two independent ones for the repeated root).
- Combine the exponential modes into the general solution.
- Verify by substitution.
Step 1 · Characteristic equation
With $A=\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$,
$$ \lambda^3-3\lambda-2=(\lambda-2)(\lambda+1)^2=0, $$
so $\lambda_1=2$ (simple) and $\lambda_2=-1$ (multiplicity 2).
Step 2 · Eigenvectors
For $\lambda_1=2$: $\mathbf v_1=(1,1,1)$. For $\lambda_2=-1$, the equation $v_1+v_2+v_3=0$ has a two-dimensional solution space, so choose
$$ \mathbf v_2=\begin{pmatrix}1\\-1\\0\end{pmatrix},\qquad \mathbf v_3=\begin{pmatrix}1\\0\\-1\end{pmatrix}. $$
Step 3 · General solution
$$ \begin{pmatrix}x\\y\\z\end{pmatrix}=C_1e^{2t}\begin{pmatrix}1\\1\\1\end{pmatrix}+C_2e^{-t}\begin{pmatrix}1\\-1\\0\end{pmatrix}+C_3e^{-t}\begin{pmatrix}1\\0\\-1\end{pmatrix}. $$
Step 4 · Verification
Substituting $x,y,z$ into all three equations $\dot x=y+z$, $\dot y=x+z$, $\dot z=x+y$ confirms each identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=2$ (simple), $\lambda_2=-1$ (multiplicity 2) |
| Geometric multiplicity of $\lambda_2$ | 2 (no generalised eigenvector needed) |
| Eigenvector for $\lambda_1$ | $(1,1,1)$ |
Additional Notes
- Because the matrix is symmetric, the repeated eigenvalue's algebraic and geometric multiplicities coincide, so three genuine eigenvectors exist.
- Any two independent vectors satisfying $v_1+v_2+v_3=0$ work for the repeated eigenvalue; the choice is not unique.
- Verified by substitution into the original system.
Result
The general solution of the system is: