Solve the system of differential equations: ddot x + dot y +…
Solve the system of differential equations: $$ \ddot x + \dot y + x = e^t,\qquad \dot x + \ddot y = 1 $$
Explanation of Method
Writing the system with the differential operator $D=d/dt$ and eliminating $y$ by applying $D$ to the first equation and subtracting the second reduces the problem to a simple $D^3x=e^t-1$, integrated three times; consistency with the original equations then constrains the constants.
- Rewrite the system using operator notation $(D^2+1)x+Dy=e^t$ and $Dx+D^2y=1$.
- Apply $D$ to the first equation and subtract the second to eliminate $y$.
- Integrate the resulting $D^3x=e^t-1$ three times.
- Recover $y$ from the second equation.
- Impose the consistency condition required by the first equation.
Step 1 · Eliminate $y$
Applying $D$ to $(D^2+1)x+Dy=e^t$ and subtracting $Dx+D^2y=1$ cancels the $D^2y$ terms and leaves
$$ D^3x=e^t-1. $$
Step 2 · Integrate for $x$
Integrating three times,
$$ x(t)=e^t-\frac{t^3}{6}+C_1t^2+C_2t+C_3. $$
Step 3 · Recover $y$
From $D^2y=1-Dx$, integrating twice gives
$$ y(t)=-e^t+\frac{t^4}{24}-\frac{C_1t^3}{3}+\frac{(1-C_2)t^2}{2}+C_4t+C_5. $$
Step 4 · Consistency condition
Substituting back into $\ddot x+\dot y+x=e^t$ forces $2C_1+C_3+C_4=0$, so $C_4=-2C_1-C_3$, reducing the independent constants to four: $C_1,C_2,C_3,C_5$.
Step 5 · Verification
With $C_4$ eliminated this way, both original equations $\dot x+\ddot y=1$ and $\ddot x+\dot y+x=e^t$ hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Elimination result | $D^3x=e^t-1$ |
| Consistency condition | $C_4=-2C_1-C_3$ |
| Independent constants | $C_1,C_2,C_3,C_5$ (order 4 system) |
Additional Notes
- Although integrating three times naively produces 5 constants, the coupled system only has order 4, so one relation among the constants is forced by substitution.
- This consistency check is essential whenever $y$ is recovered from a lower-order relation than the one used to find $x$.
- Verified by substituting both $x(t)$ and $y(t)$ into the original coupled equations.
Result
The general solution of the system is: