Solve the system of differential equations: ddot x = 2y…
Solve the system of differential equations: $$ \ddot x = 2y,\qquad \ddot y = -2x $$
Explanation of Method
Eliminating $y$ turns the system into a single fourth-order ODE, whose characteristic equation is factored with Sophie Germain's identity into two complex-conjugate quadratic factors.
- Solve the first equation for $y$ and differentiate twice.
- Substitute into the second equation to get a fourth-order ODE in $x$.
- Factor the quartic characteristic equation.
- Write the general solution for $x$ using the complex roots.
- Recover $y$ from $x$.
Step 1 · Reduce to one equation
From $y=\ddot x/2$, differentiating twice and substituting into $\ddot y=-2x$ gives
$$ x^{(4)}+4x=0. $$
Step 2 · Factor the characteristic equation
Using Sophie Germain's identity, $r^4+4=(r^2-2r+2)(r^2+2r+2)$, with roots
$$ r=1\pm i,\qquad r=-1\pm i. $$
Step 3 · General solution for $x$
$$ x(t)=e^{t}(C_1\cos t+C_2\sin t)+e^{-t}(C_3\cos t+C_4\sin t). $$
Step 4 · Recover $y$
Computing $\ddot x$ and dividing by $2$,
$$ y(t)=e^{t}(C_2\cos t-C_1\sin t)+e^{-t}(C_3\sin t-C_4\cos t). $$
Step 5 · Verification
Substituting into $\ddot x=2y$ and $\ddot y=-2x$ confirms both equations hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Reduced ODE | $x^{(4)}+4x=0$ |
| Roots | $1\pm i,\ -1\pm i$ |
| Factorisation tool | Sophie Germain's identity |
Additional Notes
- Sophie Germain's identity $r^4+4a^4=(r^2-2ar+2a^2)(r^2+2ar+2a^2)$ is the fastest route to factor $r^4+4$ without guessing roots.
- Each complex-conjugate pair $\alpha\pm\beta i$ contributes an $e^{\alpha t}(\cos\beta t,\sin\beta t)$ pair of real solutions.
- Verified by direct substitution into the original second-order system.
Result
The general solution of the system is: