Homogeneous first-order differential equation ($y^{\prime }=F(y/x)$)

A homogeneous first-order differential equation has the form $y^{\prime }=F(y/x)$ on a domain where $x\ne 0$. The substitution $v=y/x$, so $y^{\prime }=v+x{v}^{\prime }$, reduces it toWhere $F(v)-v\ne 0$, this is a separable differential equation, with $\int dv/[F(v)-v]=\ln |x|+C$. Values $v_0$ satisfying $F(v_0)=v_0$ give the additional straight-line solutions $y=v_{0}x$, which must be checked before division. This scaling-based use of “homogeneous” is different from a homogeneous linear differential equation.

For example, $y^{\prime }=y/x+x/y$ gives $F(v)=v+1/v$, so $v{v}^{\prime }=1/x$ and $y^2=x^2(2\ln |x|+C)$. Real explicit branches are $y=\pm x\sqrt{2\ln |x|+C}$ on intervals where the expression under the square root is positive. Zeros of a denominator and endpoints of a branch must still be checked in the original ordinary differential equation.

For real $a\ne 0$ and $x\ne 0$, set $u=y/x$ to obtain $x{u}^{\prime }=a\tan u$. On nonconstant branches, separation gives $\log |\sin u|=a\log |x|+C$, hence $\sin (y/x)=C_1|x{|}^a$. Zeros of the divided-out tangent give $y=k\pi x$, already represented by $C_1=0$. Points where $\cos (y/x)=0$ remain excluded because the original differential equation is undefined. An implicit trigonometric identity alone does not specify a connected branch or its domain.

For $x>0$ on a domain where the tangent is finite, put $v=y/x$. The homogeneous first-order differential equation becomes $x{v}^{\prime }=-\tan v$. Direct differentiation gives $(x\sin v{)}^{\prime }=0$, hence $x\sin (y/x)=C$, without dividing by $\sin v$ and losing constant-ratio solutions. If $C=x_0>0$ and $y/x\to \pi /2$ as $x\downarrow x_0$, two branches are $y=x\arcsin (x_0/x)$ and $y=x[\pi -\arcsin (x_0/x)]$, defined for $x>x_0$. They approach the same finite endpoint but have unbounded derivatives. The original tangent is undefined at the endpoint, so neither is a classical solution through that point. An implicit relation extending continuously to an initial point does not by itself solve an initial value problem there.

An implicit homogeneous first-order differential equation $y/x=F(p)$, where $p=dy/dx$, can be reduced by treating the slope as an auxiliary variable. Differentiating $y=xF(p)$ givesWhere the divisions are valid, this is a separable differential equation. Constant slopes lost when dividing by $p-F(p)$ must be checked in the original equation.

For $F(p)=2p/(1-p^2)$, the reduced equation is $2x{p}^{\prime }=-p(1-p^2)$ and its nonconstant solutions obey $x{p}^2/(1-p^2)=C$. Substitution recovers the parabolas $y^2=4C(x+C)$, together with the constant solution $y=0$. Real graph branches must have $p^2=C/(x+C)\ge 0$, and the original restrictions $x\ne 0$, $p\ne \pm 1$ still apply. The constant slopes $p=\pm 1$ solve the reduced equation but are excluded from the original equation.