\documentstyle[11pt]{article}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\topmargin}{-.2in}
\setlength{\textheight}{9.5in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\parindent}{0pt}
\setlength{\parskip}{5pt}

\begin{document}
\pagestyle{empty}

{\Large\bf Solution Techniques for Problems of the Form y''=f(y,y')}

\vspace{.3in}

A second-order differential equation of the form
$$
y''=f(y,y')
$$
can be converted into a first-order differential equation by
assuming y that is the independent variable and y' is the
new dependent variable. Suppose that we can write

$$
y'(t)=v(y(t))
$$

Then 

$$
y''(t)=\frac{dv}{dy}\frac{dy}{dt}=v'(y)y'(t)=v'(y)v(y)
$$

Substituting these expressions into the original differential equation
we obtain

$$
y''=v'(y)v(y)=f(y,v(y))
$$

Dividing both sides by v(y), we obtain

$$
v'(y)=\frac{f(y,v(y))}{v(y)}
$$

This is a first-order ode for v(y). Suppose we can compute v(y).
Then

$$
y'(t)=v(y)
$$

This is a separable first-order differential equation for y(t). 

\vspace{.3in}

{\bf Example}

\vspace{.2in}

Consider the ode

$$
y''=(y')^3y^2
$$

This ode is of the correct type, since y'' depends does not depend
explicitly on t. Making the substitution suggested above, we convert
the ode into

$$
v'=v^3y^2/v=v^2y^2
$$

This is a separable ode in the variable v. The solution is

$$
v=-\frac{3}{y^3+C_1}
$$

where C_1 is a constant. This leads to the new differential
equation

$$
y'=-\frac{3}{y^3+C_1}
$$

This is a separable ode. We have

$$
\int(y^3+C_1)dy=-\int 3 dt +C_2
$$

where C_2 is a second constant. Integrating both sides, we obtain 

$$
\frac{y^4}{4}+C_1y=-3t+C_2
$$

\end{document}
