Solve the system of differential equations: ddot x = 3x +…
Solve the system of differential equations: $$ \ddot x = 3x + 4\dot y,\qquad \ddot y = 3y + 4\dot x,\quad x(0)=2,\ y(0)=2,\ \dot x(0)=0,\ \dot y(0)=0 $$
Explanation of Method
Adding and subtracting the equations decouples the system into independent ODEs for $u=x+y$ and $v=x-y$; the symmetric initial data forces $v\equiv0$, meaning $x(t)=y(t)$ for all time.
- Define $u=x+y$, $v=x-y$ and add/subtract the equations.
- Solve the resulting second-order ODEs for $u$ and $v$.
- Convert the initial conditions to conditions on $u,v$.
- Solve for the constants; note the resulting symmetry.
Step 1 · Decouple
With $u=x+y$, $v=x-y$, adding and subtracting the two equations gives
$$ \ddot u-4\dot u-3u=0,\qquad \ddot v+4\dot v-3v=0. $$
Step 2 · Solve for $u$ and $v$
The characteristic roots are $r=2\pm\sqrt7$ for $u$ and $r=-2\pm\sqrt7$ for $v$:
$$ u(t)=A_1e^{(2+\sqrt7)t}+A_2e^{(2-\sqrt7)t},\qquad v(t)=B_1e^{(-2+\sqrt7)t}+B_2e^{(-2-\sqrt7)t}. $$
Step 3 · Apply initial conditions
Since $u(0)=4$, $\dot u(0)=0$, $v(0)=0$, $\dot v(0)=0$, solving gives $A_1=2-\tfrac{4}{\sqrt7}$, $A_2=2+\tfrac{4}{\sqrt7}$, and $B_1=B_2=0$, so $v(t)\equiv0$.
Step 4 · Recover $x$ and $y$
Since $v\equiv0$, $x(t)=y(t)=u(t)/2$ for all $t$:
$$ x(t)=y(t)=\left(1-\frac{2}{\sqrt7}\right)e^{(2+\sqrt7)t}+\left(1+\frac{2}{\sqrt7}\right)e^{(2-\sqrt7)t}. $$
Step 5 · Verification
All four initial conditions check out, and since $x\equiv y$, both original equations reduce consistently to $\ddot x-4\dot x-3x=0$. ✓
Summary Table
| Quantity | Value |
|---|---|
| Sum-equation roots | $r=2\pm\sqrt7$ |
| Difference-equation roots | $r=-2\pm\sqrt7$ |
| Symmetric initial data | forces $v(t)\equiv0$, i.e. $x=y$ |
Additional Notes
- The symmetric initial conditions $x(0)=y(0)$, $\dot x(0)=\dot y(0)$ exactly annihilate the antisymmetric mode $v$, collapsing the trajectory onto the line $y=x$.
- For generic (non-symmetric) initial data, $v$ would not vanish and $x,y$ would differ.
- Verified against all four initial conditions and both original equations.
Result
The general solution of the system is: