When only the ratio matters
The first two guides in this rung handed you a way to recognize an exact equation and, when it was not exact, an integrating factor to rescue it. But plenty of equations refuse both routes. The next family we tame is not about exactness at all — it is about a hidden symmetry. Some right-hand sides do not really care about x and y on their own; they care only about the ratio y/x. Picture standing at a point (x, y) and reading off the slope: if the rule gives you the same slope at every point along a ray through the origin, the equation has this special structure.
Write the equation in normal form dy/dx = F(x, y). It is called a homogeneous equation (of degree zero) when F depends on x and y only through the combination y/x — that is, F(x, y) = G(y/x) for some single-variable function G. A clean test: replace x by tx and y by ty everywhere; if F(tx, ty) = F(x, y) for every t, the t's cancelled, which is exactly the signature of dependence on the ratio alone. For example dy/dx = (x + y)/x = 1 + y/x passes instantly; so does dy/dx = (x^2 + y^2)/(x y), which equals x/y + y/x — both built purely from y/x.
The substitution that unlocks it
If the equation only sees the ratio y/x, the smart move is to make that ratio the new unknown. Set v = y/x — this is the v = y/x substitution, the centerpiece of the whole guide. Then y = v x, where v is itself a function of x. The catch that trips up beginners: y is a product of two things that both depend on x, so you must use the product rule when you differentiate. That single honest step is where all the magic happens.
Differentiating y = v x by the product rule gives dy/dx = v + x (dv/dx). Now substitute both this and y/x = v into dy/dx = G(y/x): the left side becomes v + x (dv/dx), the right side becomes G(v). Rearranging, x (dv/dx) = G(v) - v. Look at what just happened — the right side now depends only on v, and the x sits alone as a coefficient. That is precisely the shape of a separable equation in v and x. A homogeneous equation is thus a member of the family of equations reducible to separable form; the substitution is the bridge.
dy/dx = G(y/x) (homogeneous) let v = y/x so y = v x product rule: dy/dx = v + x dv/dx substitute: v + x dv/dx = G(v) rearrange: x dv/dx = G(v) - v separate: dv / (G(v) - v) = dx / x integrate both sides, then put v = y/x back
Walking one all the way through
Let us solve dy/dx = (x + y)/x from start to finish, so the recipe stops being abstract. First confirm it is homogeneous: the right side is 1 + y/x, plainly a function of the ratio alone. Now apply the substitution method in the steps below. The payoff is that a problem we could not separate in x and y becomes one we can separate in x and v.
- Substitute v = y/x and dy/dx = v + x (dv/dx). The equation 1 + y/x becomes 1 + v, so v + x (dv/dx) = 1 + v.
- Cancel the v on both sides — a small gift this example gives us — leaving x (dv/dx) = 1, a clean separable equation.
- Separate and integrate: dv = dx/x gives v = ln|x| + C. (Watch for the lost case x = 0, where the original equation is undefined anyway.)
- Undo the substitution by writing v = y/x, so y/x = ln|x| + C, hence y = x (ln|x| + C). Done.
Two habits worth burning in. First, never forget to substitute back: v was only scaffolding, and an answer left in v is not an answer in the original variables. Second, the easy cancellation of v here was luck; usually G(v) - v is a messier function and the integral dv/(G(v) - v) needs partial fractions or a substitution of its own. When that integral cannot be done in elementary terms, you are left with an implicit general solution relating v (hence y/x) to x — and that is still a complete, honest answer.
Why it works, and where it fits
The geometry explains the algebra. Because the slope at (x, y) depends only on y/x, the slope field is identical along every ray from the origin: walk outward along a ray and the little direction arrows never turn. Scaling the whole picture by a factor t maps solution curves onto solution curves. The substitution v = y/x is just choosing coordinates aligned with that symmetry — and in symmetry-aligned coordinates the equation simplifies, exactly as it must. This is the deep reason a clever substitution can reduce an equation to separable form: it trades the original variables for ones the equation treats more simply.
This is the same strategic idea you will meet again and again in this rung. The v = y/x substitution kills the ratio structure; in the next guide a different substitution flattens the nonlinearity of the Bernoulli equation into a linear one; later still, substitutions tame Riccati and Clairaut equations. The lesson is not one trick but a mindset of substitution: when an equation resists direct integration, ask what change of variable makes its hidden structure visible. Homogeneous equations are the gentlest, most geometric place to learn that habit.
Cautions, cousins, and where it shines
Honest fine print. The split dv/(G(v) - v) = dx/x divides by G(v) - v, and just like ordinary separation of variables, that division can quietly drop solutions. Wherever G(v) - v = 0, you have a constant value v = c, which means y = c x — a straight line through the origin that is a genuine solution the integration skips over. So before dividing, solve G(v) = v and write down those ray solutions by hand. It is the same discipline as hunting the lost constant solutions of a separable equation, just dressed in v.
Where does all this pay off? Homogeneous equations show up wherever only a ratio is physically meaningful — angles, aspect ratios, dilution fractions. A beautiful classic is computing orthogonal trajectories: given a family of curves (say, all circles through the origin), the curves crossing them at right angles often satisfy a homogeneous equation, solved exactly by v = y/x. That is how you derive the field lines perpendicular to a family of equipotentials. The trick is small, but it opens a real door into the geometry of curve families.