Homogeneous Differential Equation

Recall that the DEGREE of a single term is the sum of the exponents (degrees) of all the variables it contains — for example, the term 3x²y²z has degree 2+2+1 = 5. A differential equation is called HOMOGENEOUS if every term in it shares the same total degree, once dy/dx itself is counted at its natural weight in that comparison. Four illustrative examples fix this classification: x + y(dy/dx) = 0 is homogeneous of degree 1 (both x and y(dy/dx) are degree-1-equivalent terms); x³y + xy³ + x²y²(dy/dx) = 0 is homogeneous of degree 4 (each term — x³y, xy³, and x²y²(dy/dx) — carries total degree 4); but x(dy/dx) + x²y = 0 and xy(dy/dx) + y² + 2x = 0 are explicitly NOT homogeneous, since their terms mix different total degrees (the first mixes degree 0 and degree 2 contributions once dy/dx is weighted in; the second mixes degree 1, 2 and 1 in an inconsistent way) — this quick check is what a student applies before attempting the substitution below.

To solve a homogeneous differential equation, the substitution y = vx is used (or x = vy, when the equation is more naturally homogeneous in x/y, e.g. when e^(x/y) terms appear). With y = vx, differentiating gives dy/dx = v + x(dv/dx) by the product rule. Substituting both y = vx and this expression for dy/dx into the original homogeneous equation always causes the bare x's (or, when using x=vy, the bare y's) that multiply identically throughout to cancel, because of the homogeneity, leaving an equation that separates cleanly into a function of v times dv on one side and a function of x times dx on the other. That separable equation is integrated in the usual way, and finally v is replaced by y/x (or x/y) to write the solution back in terms of the original variables x and y.

Three solved examples carry out the full cycle:

(i) x²y·dx − (x³+y³)dy = 0, i.e. x²y − (x³+y³)(dy/dx) = 0. With y = vx: x²(vx) − (x³+v³x³)[v + x(dv/dx)] = 0; dividing by x³ gives v − (1+v³)v − x(1+v³)(dv/dx) = 0 [after collecting terms this becomes −x(1+v³)(dv/dx) = v⁴], i.e. [(1+v³)/v⁴]dv = −dx/x, i.e. (1/v⁴ − v³/v⁴)dv + dx/x = 0. Integrating (∫v⁻⁴dv + ∫dv/v + ∫dx/x = c₁) gives −v⁻³/3 + log v + log x = c₁, i.e. log(vx) = c₁ + v⁻³/3. Since vx = y and v⁻³ = x³/y³, this becomes log y = c₁ + x³/(3y³), i.e. 3 log y = x³/y³ + c (renaming the constant). …