Solve the system of differential equations: dot x=8y, dot…
Solve the system of differential equations: $$ \dot x=8y,\qquad \dot y=-2z,\qquad \dot z=2x+8y-2z $$
Explanation of Method
The characteristic polynomial factors as $-(\lambda+2)(\lambda^2+16)$, giving a real eigenvalue $-2$ and a purely imaginary pair $\pm4i$; the resulting oscillatory modes have constant amplitude, and $x(t)\equiv z(t)$ for every basis solution — a structural feature worth checking independently.
- Write the system in matrix form and find the characteristic polynomial.
- Identify the real eigenvalue and the imaginary conjugate pair.
- Find the real eigenvector and split the complex eigenvector into real/imaginary parts.
- Assemble and verify the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}0&8&0\\0&0&-2\\2&8&-2\end{pmatrix}$, expanding $\det(A-\lambda I)$ along column 1 gives
$$ -(\lambda+2)(\lambda^2+16)=0. $$
Eigenvalues: $\lambda_1=-2$ (real) and $\lambda_{2,3}=\pm4i$ (purely imaginary conjugate pair).
Step 2 · Eigenvector for $\lambda_1=-2$
$$ \mathbf v_1=\begin{pmatrix}-4\\1\\1\end{pmatrix},\qquad A\mathbf v_1=-2\mathbf v_1.\ \checkmark $$
Step 3 · Complex eigenvector for $\lambda_2=4i$
Solving $(A-4iI)\mathbf v=0$ gives $\mathbf v(4i)=(-2i,\,1,\,-2i)$, verified against all three rows. Splitting into real and imaginary parts,
$$ \mathbf v(4i)=\underbrace{\begin{pmatrix}0\\1\\0\end{pmatrix}}_{\mathbf a}+i\underbrace{\begin{pmatrix}-2\\0\\-2\end{pmatrix}}_{\mathbf b}. $$
Step 4 · Two real oscillatory solutions
With $\alpha=0,\ \beta=4$, the two real solutions are $\mathbf u^{(2)}=\mathbf a\cos4t-\mathbf b\sin4t$ and $\mathbf u^{(3)}=\mathbf a\sin4t+\mathbf b\cos4t$, each checked directly to satisfy $\dot{\mathbf u}=A\mathbf u$. ✓
Step 5 · Structural check and verification
Since $\dfrac{d}{dt}(x-z)=8y-(2x+8y-2z)=-2(x-z)$, all three basis solutions here happen to have $x\equiv z$ (a consequence of choosing $\mathbf a,\mathbf b$ with equal first and third components). Direct substitution confirms all three original equations hold. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=-2$ (real), $\lambda_{2,3}=\pm4i$ (imaginary) |
| Structural identity | $x(t)\equiv z(t)$ for the given basis |
| Structure | diagonalisable over $\mathbb C$ |
Additional Notes
- Because $\alpha=\operatorname{Re}(\lambda_{2,3})=0$, the oscillatory modes neither grow nor decay.
- The identity $x\equiv z$ follows from $\tfrac{d}{dt}(x-z)=-2(x-z)$ together with $x(0)=z(0)$ in the chosen basis, and gives a quick sanity check on the eigenvector algebra.
- Verified by direct substitution into all three original equations.
Result
The general solution of the system is: