Solve the system of differential equations: ddot x + x = dot…
Solve the system of differential equations: $$ \ddot x + x = \dot y,\qquad 4\dot x + 2x = \dot y + 2y,\quad x(0)=0,\ y(0)=1,\ \dot x(0)=2 $$
Explanation of Method
Eliminating $y$ and $\dot y$ between the two equations produces a single third-order ODE in $x$ with three distinct real roots; the three initial conditions then pin down the constants.
- Eliminate $\dot y$ between the two equations to relate $y$ to $x$ and its derivatives.
- Differentiate and combine to obtain a single third-order ODE in $x$.
- Solve the characteristic equation and write $x(t)$.
- Recover $y(t)$ algebraically.
- Apply the three initial conditions to fix the constants.
Step 1 · Eliminate $\dot y$
From the first equation $\dot y=\ddot x+x$; substituting into the second gives $y=2\dot x+\tfrac{x}{2}-\tfrac{\ddot x}{2}$.
Step 2 · Third-order ODE for $x$
Differentiating and eliminating $y$ entirely yields
$$ x^{(3)}-2\ddot x-\dot x+2x=0,\qquad r^3-2r^2-r+2=(r-1)(r-2)(r+1)=0. $$
Step 3 · General solution for $x$ and $y$
$$ x(t)=C_1e^{t}+C_2e^{2t}+C_3e^{-t},\qquad y(t)=2C_1e^{t}+\tfrac52C_2e^{2t}-2C_3e^{-t}. $$
Step 4 · Apply initial conditions
The conditions $x(0)=0$, $\dot x(0)=2$, $y(0)=1$ give the linear system $C_1+C_2+C_3=0$, $C_1+2C_2-C_3=2$, $2C_1+\tfrac52C_2-2C_3=1$, solved by
$$ C_1=-2,\quad C_2=2,\quad C_3=0. $$
Step 5 · Particular solution and identity check
The resulting solution also satisfies the algebraic relation $(5x-2y)^2=4(y-2x)$ identically, since $5x-2y=-2e^t$ and $y-2x=e^{2t}$.
Summary Table
| Quantity | Value |
|---|---|
| Characteristic roots | $r=1,\,2,\,-1$ |
| Constants | $C_1=-2,\ C_2=2,\ C_3=0$ |
| Algebraic identity | $(5x-2y)^2=4(y-2x)$ ✓ |
Additional Notes
- Because $C_3=0$, the $e^{-t}$ mode drops out entirely for this particular initial data.
- The extra algebraic relation is a hidden conserved quantity of the specific solution, not of the general family.
- All three initial conditions and the algebraic identity were verified by direct substitution.
Result
The general solution of the system is: