Solve the system of differential equations: (dx)/(dt) = y…
Solve the system of differential equations: $$ \frac{dx}{dt} = y,\qquad \frac{dy}{dt} = x $$
Explanation of Method
Writing the system in symmetric form $dx/y=dy/x=dt$, cross-multiplying gives a direct first integral $x^2-y^2=C_1$; combining sum and difference ratios then yields two exponential integrals that solve the system explicitly.
- Write the system in symmetric form and cross-multiply to find the algebraic first integral.
- Form the sum and difference of numerator/denominator pairs to get two exponential first integrals.
- Combine these to solve explicitly for $x(t)$ and $y(t)$.
- Interpret the algebraic first integral geometrically.
Step 1 · Algebraic first integral
From $dx/y=dy/x$, cross-multiplying gives $x\,dx=y\,dy$, so
$$ x^2-y^2=C_1. $$
Step 2 · Exponential first integrals
Combining numerators and denominators, $\dfrac{d(x+y)}{x+y}=dt$ and $\dfrac{d(x-y)}{x-y}=-dt$, giving
$$ (x+y)e^{-t}=C_2,\qquad (x-y)e^{t}=C_3. $$
Step 3 · Explicit solution
Solving the two linear relations for $x$ and $y$ (and renaming constants),
$$ x(t)=C_1e^{t}+C_2e^{-t},\qquad y(t)=C_1e^{t}-C_2e^{-t}. $$
Step 4 · Verification
Differentiating gives $\dot x=y$ and $\dot y=x$ exactly, and $x^2-y^2=4C_1C_2$ is indeed constant. ✓
Summary Table
| Quantity | Value |
|---|---|
| Algebraic first integral | $x^2-y^2=C_1$ |
| Exponential integrals | $(x+y)e^{-t}=C_2,\ (x-y)e^{t}=C_3$ |
| Phase-plane picture | family of rectangular hyperbolas |
Additional Notes
- The algebraic first integral $x^2-y^2=C_1$ is simply the product of the two exponential integrals, so only two of the three are independent.
- Geometrically, every trajectory lies on a rectangular hyperbola $x^2-y^2=\text{const}$ (or the asymptote pair $x=\pm y$ when $C_1=0$).
- Verified both by direct differentiation and by confirming the first integral is constant along solutions.
Result
The general solution of the system is: