Solve the system of differential equations: dot x=y, dot…
Solve the system of differential equations: $$ \dot x=y,\qquad \dot y=z,\qquad \dot z=x $$
Explanation of Method
Since $y=\dot x$ and $z=\ddot x$, the whole system collapses to the single third-order ODE $\dddot x=x$, whose characteristic roots are the three cube roots of unity — one real root and a complex-conjugate pair.
- Observe that $y=\dot x$ and $z=\ddot x$, reducing the system to $\dddot x=x$.
- Solve the characteristic equation $r^3=1$ for the cube roots of unity.
- Write the general solution for $x(t)$ using the real root and the complex-conjugate pair.
- Recover $y=\dot x$ and $z=\ddot x$ by differentiation.
Step 1 · Reduce to a single ODE
Since $\dot x=y$, $\dot y=z$, $\dot z=x$, differentiating twice gives $\dddot x=\ddot y=\dot z=x$, so
$$ \dddot x-x=0. $$
Step 2 · Characteristic roots
The characteristic equation $r^3-1=(r-1)(r^2+r+1)=0$ gives
$$ r_1=1,\qquad r_{2,3}=\frac{-1\pm i\sqrt3}{2}. $$
Step 3 · General solution for $x$
$$ x(t)=C_1e^{t}+e^{-t/2}\!\left(C_2\cos\tfrac{\sqrt3}{2}t+C_3\sin\tfrac{\sqrt3}{2}t\right). $$
Step 4 · Recover $y$ and $z$
Since $y=\dot x$ and $z=\ddot x$, differentiating once and twice gives the corresponding expressions for $y(t)$ and $z(t)$.
Step 5 · Verification
By construction $\dot x=y$ and $\dot y=\ddot x=z$ hold automatically; $\dot z=\dddot x=x$ holds because $x$ solves $\dddot x=x$. Direct substitution of the eigenvectors $(1,\lambda,\lambda^2)$ for each cube root of unity confirms the matrix eigenvalue relation as well. ✓
Summary Table
| Quantity | Value |
|---|---|
| Reduced ODE | $\dddot x-x=0$ |
| Characteristic roots | cube roots of unity: $1,\ \tfrac{-1\pm i\sqrt3}{2}$ |
| Key relation | $y=\dot x,\ z=\ddot x$ |
Additional Notes
- Cyclic systems like $\dot x=y,\ \dot y=z,\ \dot z=x$ always reduce directly to $x^{(n)}=x$ for an $n$-variable cycle, with characteristic roots exactly the $n$-th roots of unity.
- The real root $r=1$ gives pure exponential growth, while the complex pair (both with negative real part $-\tfrac12$) gives decaying oscillation — so the $e^t$ mode eventually dominates for generic initial data.
- Verified both by the differentiation identities $y=\dot x,\ z=\ddot x$ and by checking the matrix eigenvalue relation for each cube root of unity.
Result
The general solution of the system is: