Solve the system of differential equations: 2ddot x+2dot…
Solve the system of differential equations: $$ 2\ddot x+2\dot x+x+3\ddot y+\dot y+y=0,\qquad \ddot x+4\dot x-x+3\ddot y+2\dot y-y=0 $$
Explanation of Method
Writing the system with operator polynomials in $D$, cross-multiplying to eliminate one variable at a time reduces each of $x$ and $y$ to the same fourth-order operator equation; matching coefficients across the original equations then links the remaining constants.
- Write the two equations as operator polynomials in $x$ and $y$.
- Cross-multiply and subtract to eliminate $y$, obtaining an ODE purely in $x$.
- Repeat to eliminate $x$, obtaining the same-order ODE in $y$.
- Substitute both solutions back into one of the original equations to relate the constants.
Step 1 · Operator form
$$ (2D^2+2D+1)x+(3D^2+D+1)y=0,\qquad (D^2+4D-1)x+(3D^2+2D-1)y=0. $$
Step 2 · Eliminate $y$
Cross-multiplying by the $y$-operators and subtracting gives
$$ 3D(D^2+1)(D-1)\,x=0, $$
so the characteristic roots are $r=0,\,1,\,\pm i$ and
$$ x(t)=C_1+C_2e^{t}+C_3\cos t+C_4\sin t. $$
Step 3 · Eliminate $x$
The same elimination procedure applied to $y$ gives the identical operator equation $3D(D^2+1)(D-1)y=0$, so
$$ y(t)=K_1+K_2e^{t}+K_3\cos t+K_4\sin t. $$
Step 4 · Relate the constants
Substituting both solutions into the first original equation term-by-term forces $K_1=-C_1$, $K_2=-C_2$, and
$$ K_3=\frac{-4C_3+3C_4}{5},\qquad K_4=\frac{-3C_3-4C_4}{5}. $$
Step 5 · Verification
With these relations, both original equations are satisfied term-by-term for the constant, $e^t$, and trigonometric parts. ✓
Summary Table
| Quantity | Value |
|---|---|
| Reduced operator equation | $3D(D^2+1)(D-1)=0$ |
| Roots | $r=0,\,1,\,\pm i$ |
| Independent constants | $C_1,C_2,C_3,C_4$ (four total) |
Additional Notes
- Both $x$ and $y$ satisfy the same fourth-order operator equation because the elimination operator is the same on both sides by symmetry of the system.
- Only four of the eight naive constants are independent; the rest are determined by substitution into either original equation.
- Verified by matching coefficients of $1,\ e^t,\ \cos t,\ \sin t$ in both equations.
Result
The general solution of the system is: