# Homogeneous Differential Equations

Definition. A function $M(x,y)$ of two variables $x$ and $y$ is said to be homogeneous if for any parameter $\lambda$ $$M(\lambda x,\lambda y)=\lambda^n M(x,y)$$ The number $n$ is called the degree of the homogeneous function $M(x,y)$.

A first-order differential equation $$M(x,y)dx+N(x,y)dy=0$$ is said to be a homogeneous differential equation if both $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree.

Solving Homogeneous Differential Equations

Consider a homogeneous differential equation $$M(x,y)dx+N(x,y)dy=0$$ For the parameter $\lambda=\frac{1}{x}$ we obtain $$\label{eq:homeq}\frac{dy}{dx}=-\frac{M\left(1,\frac{y}{x}\right)}{N\left(1,\frac{y}{x}\right)}$$ Let $u=\frac{y}{x}$. Then $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u$. Since the RHS of \eqref{eq:homeq} can be considered as a function of $u$, say $f(u)$ we have $$x\frac{du}{dx}+u=f(u)$$ This is a separable equation as it can be written as $$\frac{du}{f(u)-u}=\frac{dx}{x}$$ and by integrating we find the solution $$\label{eq:homeq2}x=C\exp\left[\int\frac{du}{f(u)-u}\right]$$ Let $F(u)=\int\frac{du}{f(u)-u}$. Then \eqref{eq:homeq2} can be written as $$x=Ce^{F(u)}=Ce^{F\left(\frac{y}{x}\right)}$$

Example. Solve $(y-x)ydx+x^2dy=0$.

Solution. Let $M(x,y)=(y-x)y$ and $N(x,y)=x^2$. Notice that both $M$ and $N$ are homogeneous functions of degree 2. $$f(u)=-\frac{M(1,u)}{N(1,u)}=-u^2+u$$ and so $$\int\frac{du}{f(u)-u}=-\int\frac{du}{u^2}=\frac{1}{u}=\frac{x}{y}$$ Hence by \eqref{eq:homeq2} the solution is given by $$x=Ce^{\frac{x}{y}}. You don’t really have to remember the formula \eqref{eq:homeq2} to solve a homogeneous differential equation. In fact all you have to know is to use the substitution y=ux. dy=xdu+udx so the given differential equation can be written as u^2dx+xdu=0. By integrating we obtain$$\ln x-\frac{x}{y}=\ln C$$hence$$x=Ce^{\frac{x}{y}}$$Example. Solve 3x^2\frac{dy}{dx}=2x^2+y^2. Solution. The equation is a homogeneous equation. By the substitution y=ux the equation turns into$$\frac{dx}{3x}-\frac{du}{u^2-3u+2}=0$$By integrating the solution is given by$$y=\frac{(C\root 3\of{x}-2)x}{C\root 3\of{x}-1}

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