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{\Large\bf Homogeneous First-Order ODE}

\vspace{.2in}

A homegeneous first-order ode has the form:

$$
y'(t)=g(y/t) \hspace{.2in} y(t_0)=y_0
$$

where g(s) is a given function.

\vspace{.3in}

{\bf Solution Procedure}

\vspace{.2in}

To solve this problem, we introduce the variable
v(t)=y(t)/t and derive a new differential equation for
v(t). We have y(t)=v(t)t. Hence by the product rule,
y'(t)=tv'(t)+v(t). Substituting this into the original
ode we have

$$
y'(t)=tv'(t)+v(t)=g(y/t)=g(v)
$$

Cleaning this up, we have

$$
v'(t)=(g(v)-v)/t \hspace{.2in} (*)
$$

What have we accomplished? (*) is a separable ode! We now
use the techniques for separable ode to solve this problem.
After solving for v(t) we get y(t) using the formula
y(t)=tv(t).

\vspace{.3in}

{\bf Example}

\vspace{.2in}

ODE:

$$
y'=1+y/t+(y/t)^2
$$

Note that y'=g(y/t) where
$$
g(s)=1+s+s^2.
$$

We make the substitution v(t)=y(t)/t or y(t)=v(t)t and have
y'=v+tv'. Substituting this into the differential equation
and using the fact that v=y/t, we have

$$
v+tv'=1+v+v^2
$$

Cleaning this up and solving for v' we have

$$
v'=(1+v^2)/t
$$

Applying the method for separable ode, we obtain

$$
v(t)={\rm arctan}(\log|t|+C).
$$

The final solution is

$$
y(t)=t{\rm arctan}(\log|t|+C).
$$

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