1.2 Homogeneous ODEs

A homogeneous first-order ODE can be written in the form

equation (1.3) (1.3)

y^{\prime}=f(x,y)=F\left(\frac{y}{x}\right).

We will assume that $F(y/x)$ is a continuous function of its argument.

Idea: Define a new variable, $v(x)$ , such that

equation (1.4) (1.4)

y(x)=xv(x).

Rewriting (1.3) in terms of $v(x)$ and $x$ ,

y^{\prime}=F\left(\frac{y}{x}\right)\Longrightarrow\left(xv\right)^{\prime}=F(% v)\Longrightarrow xv^{\prime}+v=F(v).

Hence,

equation (1.5) (1.5)

xv^{\prime}=F(v)-v,

which can be solved by separating variables.

Example 1.5.

Find the general solution to $y^{\prime}=(y^{2}+2xy)/x^{2}$ . Then find the specific solution satisfying $y(1)=0.5$ .

Solution.

Rearranging this equation,

y^{\prime}=\left(\frac{y}{x}\right)^{2}+2\left(\frac{y}{x}\right)\equiv F\left% (\frac{y}{x}\right).

This is a homogeneous, first-order ODE, so we set $y(x)=xv(x)$ . Following the procedure leading to (1.5),

\displaystyle\left(xv\right)^{\prime}

\displaystyle=

\displaystyle v^{2}+2v

\displaystyle\Longrightarrow xv^{\prime}+v

\displaystyle=

\displaystyle v^{2}+2v

\displaystyle\Longrightarrow xv^{\prime}

\displaystyle=

\displaystyle v^{2}+v

\displaystyle\Longrightarrow xv^{\prime}

\displaystyle=

\displaystyle v(1+v).

Separating variables and integrating both sides,

\int\frac{{\rm d}v}{v(1+v)}=\int\frac{{\rm d}x}{x}.

Note that we have implicitly assumed that $v=y/x\neq 0$ and $v=y/x\neq-1$ . Neither of these special solutions are consistent with $y(1)=0.5$ , so this is valid.

To evaluate the integral on the left-hand side, use a partial fractions decomposition,

\frac{1}{v(1+v)}\equiv\frac{A}{v}+\frac{B}{1+v},

where $A$ and $B$ are constants. Hence,

\frac{1}{v(1+v)}=\frac{A(1+v)+Bv}{v(1+v)}=\frac{A+(A+B)v}{v(1+v)}.

For this to be true for any $v$ , we require that $A=1$ and $A+B=0\Longrightarrow B=-A=-1$ . Therefore

\frac{1}{v(1+v)}=\frac{1}{v}-\frac{1}{1+v}.

Substituting this into the separated ODE,

\displaystyle\int\left[\frac{1}{v}-\frac{1}{1+v}\right]\,\mathrm{d}v

\displaystyle=

\displaystyle\int\frac{{\rm d}x}{x}

\displaystyle\Longrightarrow\ln|v|-\ln|1+v|

\displaystyle=

\displaystyle\ln|x|+C,

where $C$ is a constant. Hence

\ln\left|\frac{v}{1+v}\right|=\ln|x|+C\Longrightarrow\frac{v}{1+v}=\pm\exp% \left[\ln|x|+C\right]=\pm\mathrm{e}^{C}|x|=Dx,

where $D$ is a new constant. Rearranging,

\frac{v}{1+v}=Dx\Longrightarrow v=Dx(1+v)\Longrightarrow v(1-Dx)=Dx% \Longrightarrow v=\frac{Dx}{1-Dx}.

Recalling that $v=y/x$ , this leads to the general solution

\frac{y}{x}=\frac{Dx}{1-Dx}\Longrightarrow y=\frac{Dx^{2}}{1-Dx}.

Now, given that $y(1)=0.5$ ,

\frac{1}{2}=\frac{D}{1-D}\Longrightarrow 1-D=2D\Longrightarrow 3D=1% \Longrightarrow D=\frac{1}{3}.

Hence,

y=\frac{\frac{1}{3}x^{2}}{1-\frac{1}{3}x}=\frac{x^{2}}{3-x},

which is the required answer.