\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 Bernoulli First-Order ODE}

\vspace{.2in}

A Bernoulli first-order ode has the form

$$
y'(t)=g(t)y(t)+h(t)[y(t)]^n \hspace{.2in} (*) 
$$

where g(t) and h(t) are given functions and n does not
equal 1.

\vspace{.3in}

{\bf Solution Procedure}

\vspace{.2in}

The idea is to convert the Bernoulli equation into a
linear ode. We make the substitution 

$$
v(t)=[y(t)]^{1-n}
$$

Differentiating this expression we have

$$
v'(t)=(1-n)[y(t)]^{-n}y'(t)
$$

Solving for y'(t), we have

$$
y'(t)=\frac{[y(t)]^n}{1-n}v'(t)
$$

Substituting this expression into the original ode (*), we have

$$
\frac{[y(t)]^n}{1-n}v'(t)=g(t)y(t)+h(t)[y(t)]^n
$$

Dividing both sides by y^n/(1-n) we have

$$
v'(t)=(1-n)g(t)[y(t]^{1-n})+(1-n)h(t)
$$

Replacing y^(1-n) by v, we obtain

$$
v'(t)=(1-n)g(t)v(t)+(1-n)h(t) \hspace{.2in} (**)
$$

This is a linear ode in v(t)! Solve for v(t). We obtain y(t) using
the formula

$$
y(t)=[v(t)]^{1/(1-n)}
$$

\vspace{.3in}

{\bf Example}

\vspace{.2in}

Consider the ode

$$
y'=5y-5ty^3
$$

This is a Bernoulli equation with n=3, g(t)=5, h(t)=-5t. We make
the substitution 

$$
v(t)=[y(t)]^{-2}
$$

Applying the chain rule, we have


$$
v'(t)=-2[y(t)]^{-3}y'(t)
$$

Solving for y'(t), we have

$$
y'(t)=-\frac{1}{2}[y(t)]^3v'(t)
$$

Substituting for y'(t) in the differential equation we have

$$
-\frac{1}{2}y^3v'=5y-5ty^3
$$

Dividing both sides by -.5y^3, we have

$$
v'=-\frac{10}{y^2}+10t
$$

Note that 1/y^2=v. Hence, we have

$$
v'=-10v+10t
$$

This is a linear ode for the dependent variable v(t). 
The solution is

$$
v(t)=Ce^{10t}+t-\frac{1}{10},
$$

where C is a constant. Substiting v=1/y^2, we have

$$
y^{-2}=Ce^{10t}+t-\frac{1}{10}
$$

\end{document}
