Solve the system of differential equations: (dy)/(dx) = 1…
Solve the system of differential equations: $$ \frac{dy}{dx} = 1 - \frac{1}{z},\qquad \frac{dz}{dx} = \frac{1}{y-x} $$
Explanation of Method
Substituting $u=y-x$ turns the system into a separable pair of equations in $u$ and $z$, which integrate directly to give both a first integral and the explicit solution.
- Substitute $u=y-x$ to simplify the first equation.
- Divide the two resulting equations to eliminate $x$ and separate variables in $u,z$.
- Integrate to get the relation between $u$ and $z$.
- Substitute back and integrate again for $u(x)$.
- Recover $y(x)$ and $z(x)$.
Step 1 · Substitution
With $u=y-x$, $\dfrac{du}{dx}=\dfrac{dy}{dx}-1=-\dfrac1z$, so the system becomes
$$ \frac{du}{dx}=-\frac1z,\qquad \frac{dz}{dx}=\frac1u. $$
Step 2 · Separate and integrate
Dividing gives $\dfrac{dz}{du}=-\dfrac{z}{u}$, i.e. $\dfrac{dz}{z}=-\dfrac{du}{u}$. Integrating,
$$ zu=C_1. $$
Step 3 · Solve for $u(x)$
Since $z=C_1/u$, the equation $\dfrac{du}{dx}=-\dfrac{u}{C_1}$ is separable, giving
$$ u=y-x=C_2e^{-x/C_1}. $$
Step 4 · General solution
$$ y(x)=x+C_2e^{-x/C_1},\qquad z(x)=\frac{C_1}{C_2}e^{x/C_1}. $$
Step 5 · Verification
Differentiating $y$ and $z$ reproduces $\dfrac{dy}{dx}=1-\dfrac1z$ and $\dfrac{dz}{dx}=\dfrac{1}{y-x}$ exactly. ✓
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u=y-x$ |
| First integral | $z(y-x)=C_1$ |
| Second integral | $(y-x)e^{x/C_1}=C_2$ |
Additional Notes
- The pair of first integrals $z(y-x)=C_1$ and $(y-x)e^{x/C_1}=C_2$ implicitly describes every solution curve.
- Both $C_1$ and $C_2$ must be nonzero for the explicit form to make sense.
- Verified by direct differentiation and back-substitution.
Result
The general solution of the system is: