Solve the system of differential equations: dot x = y…
Solve the system of differential equations: $$ \dot x = y - 2e^{t},\qquad \dot y = x + t^2 $$
Explanation of Method
Solving the first equation for $y$ and substituting into the second collapses the system into one linear second-order ODE for $x$, which is solved by standard homogeneous plus particular-solution techniques.
- Express $y$ from the first equation and differentiate.
- Substitute into the second equation to get a single ODE in $x$.
- Solve the homogeneous equation.
- Find a particular solution for each part of the forcing term.
- Recover $y$ from $x$.
Step 1 · Reduce to one equation
From the first equation, $y=\dot x+2e^{t}$, so $\dot y=\ddot x+2e^{t}$. Substituting into $\dot y=x+t^2$ gives
$$ \ddot x - x = t^2-2e^{t}. $$
Step 2 · Homogeneous solution
The characteristic equation $r^2-1=0$ gives $r=\pm1$, so $x_h=C_1e^{t}+C_2e^{-t}$.
Step 3 · Particular solution
For $t^2$, try $x_{p1}=at^2+bt+c$, giving $x_{p1}=-t^2-2$. Since $r=1$ is a root, the forcing $-2e^t$ needs $x_{p2}=Ate^{t}$, giving $A=-1$, so
$$ x_p=-t^2-2-te^{t}. $$
Step 4 · General solution for $x$
$$ x(t)=C_1e^{t}+C_2e^{-t}-te^{t}-t^2-2. $$
Step 5 · Recover $y$
Using $y=\dot x+2e^{t}$,
$$ y(t)=(C_1+1-t)e^{t}-C_2e^{-t}-2t. $$
Step 6 · Verification
Both $\dot x=y-2e^t$ and $\dot y=x+t^2$ hold identically after substitution. ✓
Summary Table
| Quantity | Value |
|---|---|
| Reduced ODE | $\ddot x-x=t^2-2e^t$ |
| Homogeneous roots | $r=\pm1$ |
| Resonant term | $-te^{t}$ (since $r=1$ is a root) |
Additional Notes
- $e^t$ is a solution of the homogeneous equation, so its particular-solution ansatz must be multiplied by $t$ to avoid duplication (resonance).
- $y$ is recovered algebraically once $x(t)$ is known — no separate integration constant is introduced for $y$.
- Verified by substitution into both original first-order equations.
Result
The general solution of the system is: