Solve the system of differential equations: dot x=2x-y+2z…
Solve the system of differential equations: $$ \dot x=2x-y+2z,\qquad \dot y=x+2z,\qquad \dot z=-2x+y-z $$
Explanation of Method
The characteristic polynomial factors as $(\lambda-1)(\lambda^2+1)$, giving one real eigenvalue and a purely imaginary conjugate pair $\pm i$; the complex eigenvector is split into real and imaginary parts to build two real oscillatory solutions.
- Write the system in matrix form and find the characteristic polynomial.
- Identify the real eigenvalue and the complex conjugate pair.
- Find the real eigenvector and the complex eigenvector.
- Split the complex eigenvector into real/imaginary parts to build two real solutions.
- Assemble and verify the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}2&-1&2\\1&0&2\\-2&1&-1\end{pmatrix}$, expanding $\det(A-\lambda I)$ gives
$$ -\lambda^3+\lambda^2-\lambda+1=-(\lambda-1)(\lambda^2+1)=0. $$
Eigenvalues: $\lambda_1=1$ (real) and $\lambda_{2,3}=\pm i$ (complex conjugate pair).
Step 2 · Eigenvector for $\lambda_1=1$
$$ \mathbf v_1=\begin{pmatrix}0\\2\\1\end{pmatrix},\qquad A\mathbf v_1=\mathbf v_1.\ \checkmark $$
Step 3 · Complex eigenvector for $\lambda_2=i$
Solving $(A-iI)\mathbf v=0$ gives $\mathbf v(i)=(-1-i,\,-1-i,\,1)$, verified against all three rows. Splitting into real and imaginary parts,
$$ \mathbf v(i)=\underbrace{\begin{pmatrix}-1\\-1\\1\end{pmatrix}}_{\mathbf a}+i\underbrace{\begin{pmatrix}-1\\-1\\0\end{pmatrix}}_{\mathbf b}. $$
Step 4 · Two real oscillatory solutions
With $\alpha=0,\ \beta=1$, the two real solutions are
$$ \mathbf u^{(2)}=\mathbf a\cos t-\mathbf b\sin t,\qquad \mathbf u^{(3)}=\mathbf a\sin t+\mathbf b\cos t, $$
each checked directly to satisfy $\dot{\mathbf u}=A\mathbf u$. ✓
Step 5 · General solution and verification
Writing $P=C_2+C_3,\ Q=C_2-C_3$ for brevity, substitution into all three original equations confirms the solution below satisfies the system identically (note $\dot z=C_1e^t-C_2\sin t+C_3\cos t$). ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=1$ (real), $\lambda_{2,3}=\pm i$ (complex) |
| Real eigenvector | $(0,2,1)$ |
| Complex eigenvector split | $\mathbf a=(-1,-1,1),\ \mathbf b=(-1,-1,0)$ |
Additional Notes
- Since $\alpha=\operatorname{Re}(\lambda_2)=0$, the oscillatory modes have constant amplitude — no growth or decay envelope.
- Verified term-by-term by matching the coefficients of $e^t,\ \cos t,\ \sin t$ in all three equations.
Result
The general solution of the system is: