Homogeneous First-Order ODE

What "homogeneous" means here

Homogeneous gets used for (at least) two completely different things in differential equations — be careful:

• Homogeneous linear ODE (right side is zero): $a_n y^{(n)} + \cdots + a_0 y = 0$. Relevant for #183–#185.
• Homogeneous first-order ODE (this problem): the right side $dy/dx = F(x, y)$ is a function of the ratio $y/x$ only: $\frac{dy}{dx} = F\!\left(\frac{y}{x}\right).$

Equivalently, numerator and denominator are both polynomial and of the same total degree so the scaling cancels. Example of degree-2/degree-2: $\frac{dy}{dx} = \frac{x^{2} + y^{2}}{x \, y}$ Every term on top and bottom is degree-2, so this is homogeneous.

Why the substitution $y = v x$ works

Let $v = y/x$ (so $v$ depends on $x$). Then $y = v x$ and, by product rule, $\frac{dy}{dx} = v + x \, \frac{dv}{dx}.$

Substituting into $dy/dx = F(y/x) = F(v)$: $v + x \, v' = F(v) \implies x \, v' = F(v) - v$ $\frac{dv}{F(v) - v} = \frac{dx}{x}$

That is separable. Integrate, then back-substitute $v = y/x$.

The given equation

$\frac{dy}{dx} = \frac{x^{2} + y^{2}}{x y}$

Let's verify homogeneity first. Divide top and bottom by $x^{2}$: $\frac{dy}{dx} = \frac{1 + (y/x)^{2}}{(y/x)} = \frac{1 + v^{2}}{v}.$

So $F(v) = (1 + v^{2})/v$. Homogeneous. ✓

Step-by-step solution

Step 1 — Substitute $y = v x$.

$v + x \, v' = \frac{1 + v^{2}}{v}$

Step 2 — Isolate $x v'$. $x \, v' = \frac{1 + v^{2}}{v} - v = \frac{1 + v^{2} - v^{2}}{v} = \frac{1}{v}$

Step 3 — Separate. $v \, dv = \frac{dx}{x}$

Step 4 — Integrate. $\int v \, dv = \int \frac{dx}{x}$ $\frac{v^{2}}{2} = \ln|x| + C_1$ $v^{2} = 2 \ln|x| + C \quad (C = 2 C_1)$

Step 5 — Back-substitute $v = y/x$. $\left(\frac{y}{x}\right)^{2} = 2 \ln|x| + C$ $\boxed{\, y^{2} = x^{2} \bigl(2 \ln|x| + C\bigr) \,}$

(For $x > 0$, $|x| = x$.)

Verification

Implicit differentiation of $y^{2} = 2 x^{2} \ln x + C x^{2}$: $2 y \, y' = 4 x \ln x + 2 x + 2 C x = 2 x (2 \ln x + 1 + C)$ $y \, y' = x (2 \ln x + 1 + C)$

Also, $y^{2} / x^{2} = 2 \ln x + C$, so $2 \ln x + C = y^{2}/x^{2}$. Plug back: $y \, y' = x \!\left( \frac{y^{2}}{x^{2}} + 1 \right) = \frac{y^{2}}{x} + x = \frac{y^{2} + x^{2}}{x}$

Divide by $y$: $y' = \frac{y^{2} + x^{2}}{x \, y} \quad\checkmark$

Geometric picture

The direction field of a homogeneous ODE is radial-symmetric in a scaling sense: the slope at $(x, y)$ depends only on the angle $\theta = \arctan(y/x)$, not on how far out from the origin you are. Along any ray through the origin the slope is constant. That's exactly why the substitution $y = v x$ (moving along rays) strips the radial dependence away.

Quick detection checklist

Is your equation $dy/dx = f(x, y)$ homogeneous? Two quick tests:

1. Scaling test. Replace $(x, y) \to (t x, t y)$. If $f(tx, ty) = f(x, y)$ for all $t \ne 0$, it's homogeneous (degree-0).
2. Ratio test. Write $f$ as a single fraction. If every term in the numerator has the same degree $d$, and likewise for the denominator, and the two degrees match, it's homogeneous.

Common mistakes

• Confusing the two meanings of "homogeneous" — this one is about the right-side scaling, not about the equation being set to zero.
• Forgetting the $v$ in $dy/dx = v + x \, dv/dx$. Students often write $dy/dx = x \, dv/dx$ only.
• Using $y/x$ as a free variable without explicitly substituting. Define $v$ up front and track it carefully.
• Absolute values. $\ln|x|$ matters when $x < 0$ — pick a domain and stay on one side of zero.

When the substitution stalls

If the ODE isn't quite homogeneous but has a linear-shift form like $\frac{dy}{dx} = \frac{a x + b y + c}{d x + e y + f}, \qquad c, f \ne 0,$ you can often translate $(x, y) \to (x - x_0, y - y_0)$ to kill $c, f$ first, then apply the homogeneous substitution. Or: if the two lines $ax + by + c$, $dx + ey + f$ are parallel, use the substitution $w = a x + b y$ (a different linearization).

Try it in the visualization

Slide the initial condition along a ray through the origin and watch the slope stay the same (the hallmark of a homogeneous ODE). Overlay the substitution $y = v x$ — the map that sends rays through the origin into vertical lines in the $(x, v)$ plane, turning the radial ODE into a separable one.