*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 \begin{equation}\label{eq:homeq}\frac{dy}{dx}=-\frac{M\left(1,\frac{y}{x}\right)}{N\left(1,\frac{y}{x}\right)}\end{equation} 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 \begin{equation}\label{eq:homeq2}x=C\exp\left[\int\frac{du}{f(u)-u}\right]\end{equation} 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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