Solve the system of differential equations: 5ddot x+5x+3ddot…
Solve the system of differential equations: $$ 5\ddot x+5x+3\ddot y+3y+8\ddot z+8z=0,\quad \ddot x+4x+3\ddot y+12y+3\ddot z+12z=0,\quad \ddot x-2x-2\ddot y-11y-9z=0 $$
Explanation of Method
The first two equations factor neatly as $(D^2+1)$ and $(D^2+4)$ applied to linear combinations of $x,y,z$; combining those first integrals with the third equation isolates a single ODE for $x$, from which $y$ and $z$ follow algebraically.
- Recognise that equations (1) and (2) are $(D^2+1)$ and $(D^2+4)$ applied to specific linear combinations.
- Integrate each to get two intermediate relations involving sinusoids.
- Combine these relations with the third equation to isolate an ODE for $x$ alone.
- Recover $y$ and $z$ from the intermediate relations.
Step 1 · Group the first two equations
$$ (D^2+1)(5x+3y+8z)=0,\qquad (D^2+4)(x+3y+3z)=0. $$
Integrating each gives
$$ 5x+3y+8z=A_1\cos t+A_2\sin t,\qquad x+3y+3z=B_1\cos 2t+B_2\sin 2t. $$
Step 2 · Combine with the third equation
Eliminating $y$ and $z$ between these two relations and equation (3), and matching like terms, reduces to
$$ \ddot x+x=-3(B_1\cos 2t+B_2\sin 2t). $$
Step 3 · Solve for $x$
The homogeneous part gives $\cos t,\sin t$; a $\cos2t,\sin2t$ particular ansatz gives coefficients $B_1,B_2$ directly, so
$$ x(t)=C_1\cos t+C_2\sin t+C_5\cos 2t+C_6\sin 2t. $$
Step 4 · Recover $y$ and $z$
Substituting back into the two intermediate relations gives
$$ y(t)=\frac{7C_1-3C_3}{15}\cos t+\frac{7C_2-3C_4}{15}\sin t+C_5\cos 2t+C_6\sin 2t, $$
$$ z(t)=\frac{-4C_1+C_3}{5}\cos t+\frac{-4C_2+C_4}{5}\sin t-C_5\cos 2t-C_6\sin 2t. $$
Step 5 · Verification
Matching the $\cos t,\sin t,\cos 2t,\sin 2t$ coefficients confirms all three original equations hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Grouped combination 1 | $(D^2+1)(5x+3y+8z)=0$ |
| Grouped combination 2 | $(D^2+4)(x+3y+3z)=0$ |
| Reduced ODE for $x$ | $\ddot x+x=-3(B_1\cos2t+B_2\sin2t)$ |
Additional Notes
- Recognising the special structure in equations (1) and (2) — where the coefficients of $\ddot{\,}$ and the undifferentiated variable are proportional — is what makes this $3\times3$ system tractable without a full eigenvalue analysis.
- The system has order $2\times3=6$, matching the six independent constants $C_1,\dots,C_6$ in the final solution.
- Verified by matching coefficients of each independent function ($1,\cos t,\sin t,\cos 2t,\sin 2t$) across all three equations.
Result
The general solution of the system is: