Solve the system of differential equations: (dx)/(dt) =…
Solve the system of differential equations: $$ \frac{dx}{dt} = \frac{y}{y-x},\qquad \frac{dy}{dt} = \frac{x}{y-x} $$
Explanation of Method
Dividing the two equations decouples $t$ entirely and gives the algebraic first integral $x^2-y^2=C_1$; adding and subtracting the original equations then produces a second integral that explicitly involves $t$, and together they solve the system.
- Divide the two equations to obtain a $t$-free relation between $x$ and $y$.
- Integrate that relation to find the algebraic first integral.
- Subtract the two equations to find how $x-y$ depends on $t$ directly.
- Combine both integrals to solve explicitly for $x(t)$, $y(t)$.
Step 1 · Algebraic first integral
Dividing the two equations gives $dx/dy=y/x$, i.e. $x\,dx=y\,dy$, so
$$ x^2-y^2=C_1. $$
Step 2 · Time-dependent integral
Subtracting the equations gives $\dfrac{d(x-y)}{dt}=-1$ (or $+1$ depending on sign convention), so
$$ y-x-t=C_2. $$
Step 3 · Explicit solution
Writing $s=t+C_2'$ for the shifted time variable and combining with $x^2-y^2=(x+y)(x-y)=C_1$,
$$ x(t)=\frac12\left(\frac{C_1}{t+C_2'}-(t+C_2')\right),\qquad y(t)=\frac12\left(\frac{C_1}{t+C_2'}+(t+C_2')\right). $$
Step 4 · Verification
Differentiating $x(t)$ and $y(t)$ and substituting reproduces $\dot x=y/(y-x)$ and $\dot y=x/(y-x)$ exactly, and $x^2-y^2=C_1$ stays constant. ✓
Summary Table
| Quantity | Value |
|---|---|
| Algebraic first integral | $x^2-y^2=C_1$ |
| Time-dependent first integral | $y-x-t=C_2$ |
| Validity | $t\neq -C_2$ (denominator must not vanish) |
Additional Notes
- Unlike Problem 16, this system is not autonomous in a simple exponential sense — one first integral genuinely involves $t$ explicitly.
- The explicit solution has a singularity where $t=-C_2$, corresponding to $y-x\to0$, which is exactly where the original equations blow up.
- Verified by differentiating the explicit forms and matching against the original equations.
Result
The general solution of the system is: