Solve the system of differential equations: 6ddot x-3dot…
Solve the system of differential equations: $$ 6\ddot x-3\dot x+6x+\dot y=t,\qquad \dot x+8x+y=0,\quad x(0)=\tfrac{11}{36},\ y(0)=-\tfrac{22}{9},\ \dot x(0)=\tfrac16 $$
Explanation of Method
Solving the second equation for $y$ and $\dot y$ reduces the system to a single second-order linear ODE for $x$ with polynomial forcing; applying the initial conditions collapses the homogeneous part entirely, leaving a solution that traces a straight line in the $(x,y)$-plane.
- Solve the second equation for $y$ and differentiate to get $\dot y$.
- Substitute into the first equation to obtain a second-order ODE in $x$.
- Solve the homogeneous and particular parts.
- Recover $y(t)$ and apply the initial conditions.
- Eliminate $t$ between $x$ and $y$ to find the line the trajectory lies on.
Step 1 · Reduce to one equation
From $y=-\dot x-8x$, differentiating and substituting into the first equation gives
$$ 5\ddot x-11\dot x+6x=t. $$
Step 2 · Homogeneous and particular solutions
The characteristic equation $5r^2-11r+6=0$ gives $r=\tfrac65,\,1$. A linear ansatz $x_p=at+b$ gives $a=\tfrac16$, $b=\tfrac{11}{36}$, so
$$ x(t)=C_1e^{6t/5}+C_2e^{t}+\frac{t}{6}+\frac{11}{36}. $$
Step 3 · Apply initial conditions
The conditions $x(0)=\tfrac{11}{36}$ and $\dot x(0)=\tfrac16$ force $C_1=C_2=0$, so the homogeneous part vanishes entirely and
$$ x(t)=\frac{t}{6}+\frac{11}{36},\qquad y(t)=-\frac{4t}{3}-\frac{47}{18}. $$
Step 4 · The trajectory is a straight line
Eliminating $t=6x-\tfrac{11}{6}$ and substituting into $y(t)$ gives
$$ y=-8x-\frac16, $$
confirming that every point $(x,y)$ on the solution lies on this line.
Step 5 · Verification
Substituting back confirms $5\ddot x-11\dot x+6x=t$ and $\dot x+8x+y=0$ hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Reduced ODE | $5\ddot x-11\dot x+6x=t$ |
| Homogeneous roots | $r=\tfrac65,\,1$ |
| Constants from initial data | $C_1=C_2=0$ |
| Trajectory | $y=-8x-\tfrac16$ |
Additional Notes
- The given initial data happens to match the particular solution exactly, so both homogeneous constants vanish — a special, non-generic case.
- Because $x$ is a linear function of $t$, eliminating $t$ between $x(t)$ and $y(t)$ always produces a straight line for this initial-value problem.
- Both the differential equations and the line relation were verified by direct substitution.
Result
The general solution of the system is: