# 3.5: Separable Differential Equations

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A first order differential equation is {\color{blue}\it

separable} if it can be written as

\begin{equation} \label{eq:2.2.1}

h(y)y'=g(x),

\end{equation}

where the left side is a product of $y'$ and a function of $y$ and

the right side is a function of $x$. Rewriting a separable

differential equation in this form is called {\color{blue}\it separation of

variables.\/} In Section~2.1 we used separation of variables

to solve homogeneous linear equations. In this section we'll apply

this method to nonlinear equations.

To see how to solve \eqref{eq:2.2.1}, let's first assume that $y$

is a solution.

Let $G(x)$ and $H(y)$ be antiderivatives of $g(x)$

and $h(y)$; that is,

\begin{equation} \label{eq:2.2.2}

H'(y)=h(y)\mbox{\quad and \quad}G'(x)=g(x).

\end{equation}

Then, from the chain rule,

$$

{d\over dx}H(y(x))=H'(y(x))y'(x)=h(y)y'(x).

$$

Therefore \eqref{eq:2.2.1} is equivalent to

$$

{d\over dx}H(y(x))={d\over dx}G(x).

$$

Integrating both sides of this equation and combining the constants of

integration yields

\begin{equation} \label{eq:2.2.3}

H(y(x))=G(x)+c.

\end{equation}

Although we derived this equation on the assumption that $y$

is a solution of \eqref{eq:2.2.1}, we can now view it differently:

Any differentiable function $y$ that satisfies \eqref{eq:2.2.3}

for some constant $c$ is a solution of \eqref{eq:2.2.1}. To see this, we

differentiate both sides of \eqref{eq:2.2.3}, using the chain rule on the

left, to obtain

$$

H'(y(x))y'(x)=G'(x),

$$

which is equivalent to

$$

h(y(x))y'(x)=g(x)

$$

because of \eqref{eq:2.2.2}.

In conclusion, to solve \eqref{eq:2.2.1} it suffices to find

functions $G=G(x)$ and $H=H(y)$ that satisfy \eqref{eq:2.2.2}.

Then any differentiable function $y=y(x)$ that satisfies

\eqref{eq:2.2.3} is a solution of \eqref{eq:2.2.1}.

\begin{example}\label{example:2.2.1}\rm

Solve the equation

$$

y'=x(1+y^2).

$$

\end{example}

\solution Separating variables

yields

$$

{y'\over 1+y^2}=x.

$$

Integrating yields

$$

\tan^{-1}y={x^2\over2}+c

$$

Therefore

$$

y=\tan\left({x^2\over2}+c\right).

$$

\begin{example}\label{example:2.2.2} \rm \mbox{}\newline

\begin{alist}

\item %(a)

Solve the equation

\begin{equation} \label{eq:2.2.4}

y'=-{x\over y}.

\end{equation}

\item %(b)

Solve the initial value problem

\begin{equation} \label{eq:2.2.5}

y'=-{x\over y}, \quad y(1)=1.

\end{equation}

\item %(c)

Solve the initial value problem

\begin{equation} \label{eq:2.2.6}

y'=-{x\over y}, \quad y(1)=-2.

\end{equation}

\end{alist}

\end{example}

\solutionpart{a} Separating variables

in \eqref{eq:2.2.4} yields

$$

yy'=-x.

$$

Integrating yields

$$

{y^2\over2}=-{x^2\over2}+c,\mbox{\quad or,\, equivalently, \quad}

x^2+y^2=2c.

$$

The last equation shows that $c$ must be positive if $y$ is to be a

solution of \eqref{eq:2.2.4} on an open interval. Therefore we let

$2c=a^2$ (with $a > 0$) and rewrite the last equation as

\begin{equation} \label{eq:2.2.7}

x^2+y^2=a^2.

\end{equation}

This equation has two differentiable solutions for $y$ in terms of

$x$:

\begin{equation} \label{eq:2.2.8}

y=\phantom{-} \sqrt{a^2-x^2}, \quad -a < x < a,

\end{equation}

and

\begin{equation} \label{eq:2.2.9}

y= - \sqrt{a^2-x^2}, \quad -a < x < a.

\end{equation}

The solution curves defined by \eqref{eq:2.2.8} are semicircles

above the

$x$-axis and those defined by \eqref{eq:2.2.9} are semicircles below the

$x$-axis (Figure~\ref{figure:2.2.1}).

\solutionpart{b} The solution of \eqref{eq:2.2.5}

is positive when $x=1$; hence, it is of the form \eqref{eq:2.2.8}.

Substituting $x=1$ and $y=1$ into \eqref{eq:2.2.7} to satisfy the

initial condition yields $a^2=2$; hence, the solution of

\eqref{eq:2.2.5}

is

$$

y=\sqrt{2-x^2}, \quad - \sqrt{2}< x < \sqrt{2}.

$$

\solutionpart{c} The solution of \eqref{eq:2.2.6}

is negative when $x=1$ and is therefore of the form \eqref{eq:2.2.9}.

Substituting $x=1$ and $y=-2$ into \eqref{eq:2.2.7} to satisfy the

initial condition yields $a^2=5$. Hence, the solution of \eqref{eq:2.2.6}

is

$$

y=- \sqrt{5-x^2}, \quad -\sqrt{5} < x < \sqrt{5}.

$$

\begin{figure}[H]

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020201}

\color{blue}

\caption{

{\bf (a)} $y=\sqrt{2-x^{2}}$,\, $-\sqrt{2}<x<\sqrt{2}$; \,

{\bf (b)} $y=-\sqrt{5-x^{2}}$,\, $-\sqrt{5}<x<\sqrt{5}$}

\label{figure:2.2.1}

\end{figure}

\boxit{Implicit Solutions of Separable Equations}

\noindent

In Examples~\ref{example:2.2.1} and \ref{example:2.2.2} we were able to

solve the equation $H(y)=G(x)+c$ to obtain explicit formulas for

solutions of the given separable differential equations. As we'll

see in the next example, this isn't always possible.

In this situation we must broaden our definition of a solution of a

separable equation. The next theorem provides the basis for this

modification. We omit the proof, which requires a result from

advanced calculus called as the {\color{blue}\it implicit function

theorem\/}.

\begin{theorem}\color{blue}~\label{thmtype:2.2.1}

Suppose $g=g(x)$ is continous on $(a,b)$ and $h=h(y)$

are continuous on $(c,d).$ Let $G$ be an antiderivative of

$g$ on $(a,b)$ and let $H$ be an antiderivative of $h$ on $(c,d).$

Let $x_0$ be an arbitrary point in $(a,b),$ let $y_0$

be a point in $(c,d)$ such that $h(y_0)\ne0,$ and define

\begin{equation} \label{eq:2.2.10}

c=H(y_0)-G(x_0).

\end{equation}

Then there's a function $y=y(x)$ defined on some open interval

$(a_1,b_1),$ where $a\le a_1<x_0<b_1\le b,$ such that $y(x_0)=y_0$

and

\begin{equation} \label{eq:2.2.11}

H(y)=G(x)+c

\end{equation}

for $a_1<x<b_1$.

Therefore $y$ is a solution of the initial value problem

\begin{equation} \label{eq:2.2.12}

h(y)y'=g(x),\quad y(x_0)=x_0.

\end{equation}

\end{theorem}

It's convenient to say that \eqref{eq:2.2.11} with $c$ arbitrary is an

{\color{blue}\it implicit solution\/} of $h(y)y'=g(x)$. Curves defined by

\eqref{eq:2.2.11} are integral curves of $h(y)y'=g(x)$. If $c$ satisfies

\eqref{eq:2.2.10}, we'll say that \eqref{eq:2.2.11} is an {\color{blue}\it

implicit

solution of the initial value problem\/} \eqref{eq:2.2.12}. However, keep

these points in mind:

\begin{itemize}

\item For some choices of $c$ there may not be any differentiable

functions $y$ that satisfy \eqref{eq:2.2.11}.

\item The function $y$ in \eqref{eq:2.2.11} (not \eqref{eq:2.2.11} itself)

is a solution of $h(y)y'=g(x)$.

\end{itemize}

\begin{example}\label{example:2.2.3} \rm \mbox{}\newline

\begin{alist}

\item % (a)

Find implicit solutions of

\begin{equation} \label{eq:2.2.13}

y'={2x+1\over5y^4+1}.

\end{equation}

\item % (b)

Find an implicit solution of

\begin{equation} \label{eq:2.2.14}

y'={2x+1\over5y^4+1},\quad y(2)=1.

\end{equation}

\end{alist}

\end{example}

\solutionpart{a}

Separating variables yields

$$

(5y^4+1)y'=2x+1.

$$

Integrating yields the implicit solution

\begin{equation} \label{eq:2.2.15}

y^5+y=x^2+x+ c.

\end{equation}

of \eqref{eq:2.2.13}.

\solutionpart{b} Imposing the initial condition $y(2)=1$ in

\eqref{eq:2.2.15} yields $1+1=4+2+c$, so $c=-4$. Therefore

$$

y^5+y=x^2+x-4

$$

is an implicit solution of the initial value problem \eqref{eq:2.2.14}.

Although more than one differentiable function $y=y(x)$ satisfies

\ref{eq:2.2.13}) near $x=1$, it can be shown that there's only

one such function that satisfies the initial condition $y(1)=2$.

Figure~\ref{figure:2.2.2} shows a direction field and some integral curves

for \eqref{eq:2.2.13}.

\begin{figure}[tbp]

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020202}

\color{blue}

\caption{A direction field and integral curves for

$y'=\dst{\frac{2x+1}{5y^{4}+1}}$}

\label{figure:2.2.2}

\end{figure}

\boxit{Constant Solutions of Separable Equations}

\noindent

An equation of the form

$$

y'=g(x)p(y)

$$

is separable, since it can be rewritten as

$$

{1\over p(y)}y'=g(x).

$$

However, the division by $p(y)$

is not legitimate if $p(y)=0$ for some values of $y$. The

next two examples show how to deal with this problem.

\begin{example}\label{example:2.2.4}

\rm Find all solutions of

\begin{equation} \label{eq:2.2.16}

y'=2xy^2.

\end{equation}

\end{example}

\solution Here we must divide by $p(y)=y^2$ to separate variables.

This isn't legitimate if $y$ is a solution of \eqref{eq:2.2.16} that

equals zero for some value of $x$. One such solution can be found by

inspection: $y \equiv 0$. Now suppose $y$ is a solution

of \eqref{eq:2.2.16} that isn't identically zero. Since $y$ is continuous

there must be an interval on which $y$ is never zero. Since division

by $y^2$ is legitimate for $x$ in this interval, we can separate

variables in \eqref{eq:2.2.16} to obtain

$$

{y'\over y^2}=2x.

$$

Integrating this yields

$$

-{1\over y}=x^2+c,

$$

which is equivalent to

\begin{equation} \label{eq:2.2.17}

y=-{1\over x^2+c}.

\end{equation}

We've now shown that if $y$ is a solution of \eqref{eq:2.2.16} that is

not identically zero, then $y$ must be of the form \eqref{eq:2.2.17}. By

substituting \eqref{eq:2.2.17} into \eqref{eq:2.2.16}, you can verify that

\eqref{eq:2.2.17} is a solution of \eqref{eq:2.2.16}. Thus,

solutions of \eqref{eq:2.2.16} are $y\equiv0$ and the functions

of the form \eqref{eq:2.2.17}. Note that the solution $y\equiv0$ isn't of

the form \eqref{eq:2.2.17} for any value of $c$.

Figure~\ref{figure:2.2.3} shows a direction field and some integral

curves for \eqref{eq:2.2.16}

\begin{figure}[tbp]

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020203}

\color{blue}

\caption{A direction field and integral curves for $y'=2xy^{2}$}

\label{figure:2.2.3}

\end{figure}

\begin{example}\label{example:2.2.5}

\rm Find all solutions of

\begin{equation} \label{eq:2.2.18}

y'={1\over2}x(1-y^2).

\end{equation}

\end{example}

\solution Here we must divide by $p(y)=1-y^2$ to separate variables.

This isn't legitimate if $y$ is a solution of \eqref{eq:2.2.18} that

equals $\pm1$ for some value of $x$. Two such solutions can be found

by inspection: $y \equiv 1$ and $y\equiv-1$. Now suppose

$y$ is a solution of \eqref{eq:2.2.18} such that $1-y^2$ isn't

identically zero. Since $1-y^2$ is continuous there must be an

interval on which $1-y^2$ is never zero. Since division by $1-y^2$ is

legitimate for $x$ in this interval, we can separate variables in

\eqref{eq:2.2.18} to obtain

$$

{2y'\over y^2-1}=-x.

$$

A partial fraction expansion on the left yields

$$

\left[{1\over y-1}-{1\over y+1}\right]y'=-x,

$$

and integrating yields

$$

\ln\left|{y-1\over y+1}\right|=-{x^2\over2}+k;

$$

hence,

$$

\left|{y-1\over y+1}\right|=e^ke^{-x^2/2}.

$$

Since $y(x)\ne\pm1$ for $x$ on the interval under discussion, the

quantity

$(y-1)/(y+1)$ can't change sign in this interval. Therefore

we can rewrite the last equation as

$$

{y-1\over y+1}=ce^{-x^2/2},

$$

where $c=\pm e^k$, depending upon the sign of $(y-1)/(y+1)$ on the

interval.

Solving for $y$ yields

\begin{equation} \label{eq:2.2.19}

y={1+ce^{-x^2/2}\over 1-ce^{-x^2/2}}.

\end{equation}

We've now shown that if $y$ is a solution of \eqref{eq:2.2.18} that is

not identically equal to $\pm1$, then $y$ must be as in

\eqref{eq:2.2.19}. By substituting \eqref{eq:2.2.19} into \eqref{eq:2.2.18} you

can verify that \eqref{eq:2.2.19} is a solution of

\eqref{eq:2.2.18}. Thus, the solutions of \eqref{eq:2.2.18}

are $y\equiv1$, $y\equiv-1$ and the functions of the form

\eqref{eq:2.2.19}. Note that the constant solution $y \equiv 1$ can be

obtained from this formula by taking $c=0$; however, the other

constant solution, $y \equiv -1$, can't be obtained in this way.

Figure~\ref{figure:2.2.4} shows a direction field and some integrals for

\eqref{eq:2.2.18}.

\begin{figure}[H]

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020204}

\color{blue}

\caption{

A direction field and integral curves for

$y'=\dst{\frac{x(1-y^2)}{2}}$}

\label{figure:2.2.4}

\end{figure}

\boxit{Differences Between Linear and Nonlinear Equations}

\noindent

Theorem~\ref{thmtype:2.1.2} states that if $p$ and $f$ are continuous on

$(a,b)$ then every solution of

$$

y'+p(x)y=f(x)

$$

on $(a,b)$ can be obtained by choosing a value for the constant

$c$ in the general solution, and if

$x_0$ is any point in $(a,b)$ and $y_0$ is arbitrary, then the

initial value problem

$$

y'+p(x)y=f(x),\quad y(x_0)=y_0

$$

has a solution on $(a,b)$.

The not true for nonlinear equations. First, we

saw in Examples~\ref{example:2.2.4} and \ref{example:2.2.5} that a

nonlinear

equation may have solutions that can't be obtained by choosing a

specific value of a constant appearing in a one-parameter family of

solutions. Second, it is in general impossible to determine the

interval of validity of a solution to an initial value problem for a

nonlinear equation by simply examining the equation, since the

interval of validity may depend on the initial condition. For

instance, in Example~\ref{example:2.2.2} we saw that the solution of

$$

{dy\over dx}=-{x\over y},\quad y(x_0)=y_0

$$

is valid on $(-a,a)$, where $a=\sqrt{x_0^2+y_0^2}$.

\begin{example}\label{example:2.2.6}

\rm Solve the initial value problem

$$

y'=2xy^2, \quad y(0)=y_0

$$

and determine the interval of validity of the solution.

\end{example}

\solution First suppose $y_0\ne0$. From

Example~\ref{example:2.2.4}, we know that $y$ must be of the form

\begin{equation} \label{eq:2.2.20}

y=-{1\over x^2+c}.

\end{equation}

Imposing the initial condition shows that $c=-1/y_0$.

Substituting this into \eqref{eq:2.2.20} and rearranging terms yields

the solution

$$

y= {y_0\over 1-y_0x^2}.

$$

This is also the solution if $y_0=0$. If $y_0<0$,

the denominator isn't zero for any value of $x$, so the

the solution is valid on $(-\infty,\infty)$. If $y_0>0$, the solution

is valid only on $(-1/\sqrt{y_0},1/\sqrt{y_0})$.

\exercises

In Exercises \ref{exer:2.2.1}--\ref{exer:2.2.6} find all solutions.

\begin{exerciselist}

\begin{tabular}[t]{

@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.2.1} $\dst{y'={3x^2+2x+1\over y-2}}$&

\item\label{exer:2.2.2}\vspace*{18pt} $(\sin x)(\sin y)+(\cos y)y'=0$

\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.2.3} $xy'+y^2+y=0$

& \item\label{exer:2.2.4} $y' \ln |y|+x^2y= 0$

\end{tabular}

\item\label{exer:2.2.5}

$\dst{(3y^3+3y \cos y+1)y'+{(2x+1)y\over 1+x^2}=0}$

\item\label{exer:2.2.6} $x^2yy'=(y^2-1)^{3/2}$

\exercisetext{In Exercises \ref{exer:2.2.7}--\ref{exer:2.2.10} find all

solutions. Also, plot a direction field and some integral curves on

the indicated rectangular region.}

\item\label{exer:2.2.7} \CGex $\dst{y'=x^2(1+y^2)}; \; \{-1\le x\le1,\

-1\le y\le1\}$

\item\label{exer:2.2.8} \CGex$ y'(1+x^2)+xy=0 ; \; \{-2\le x\le2,\ -1\le

y\le1\}$

\item\label{exer:2.2.9} \CGex $y'=(x-1)(y-1)(y-2); \; \{-2\le x\le2,\

-3\le y\le3\}$

\item\label{exer:2.2.10} \CGex $(y-1)^2y'=2x+3; \; \{-2\le x\le2,\ -2\le

y\le5\}$

\exercisetext{In Exercises \ref{exer:2.2.11} and \ref{exer:2.2.12} solve

the initial value problem.}

\item\label{exer:2.2.11}

$\dst{y'={x^2+3x+2\over y-2}, \quad y(1)=4}$

\item\label{exer:2.2.12} $y'+x(y^2+y)=0, \quad y(2)=1$

\exercisetext{In Exercises \ref{exer:2.2.13}-\ref{exer:2.2.16} solve

the initial value problem and graph the solution.}

\item\label{exer:2.2.13} \CGex

$(3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1$

\item\label{exer:2.2.14} \CGex

$\dst{y'+{(y+1)(y-1)(y-2)\over x+1}=0, \quad y(1)=0}$

\item\label{exer:2.2.15} \CGex $y'+2x(y+1)=0, \quad y(0)=2$

\item\label{exer:2.2.16} \CGex

$y'=2xy(1+y^2),\quad y(0)=1$

\exercisetext{In Exercises \ref{exer:2.2.17}--\ref{exer:2.2.23} solve the initial

value problem and find the interval of validity of the solution.}

\item\label{exer:2.2.17}

$y'(x^2+2)+ 4x(y^2+2y+1)=0, \quad y(1)=-1$

\item\label{exer:2.2.18}

$y'=-2x(y^2-3y+2), \quad y(0)=3$

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.2.19} $\dst{y'={2x\over 1+2y},

\quad y(2)=0}$ & \item \vspace*{5.5pt}\label{exer:2.2.20} $y'=2y-y^2,

\quad y(0)=1$ \end{tabular}

\item\label{exer:2.2.21}

$x+yy'=0, \quad y(3) =-4$

\item\label{exer:2.2.22}

$y'+x^2(y+1)(y-2)^2=0, \quad y(4)=2$

\item\label{exer:2.2.23}

$(x+1)(x-2)y'+y=0, \quad y(1)=-3$

\item\label{exer:2.2.24} Solve $\dst{y'={(1+y^2)

\over (1+x^2)}}$ explicitly.

\hint{Use the identity

$\dst{\tan(A+B)={\tan A+\tan B\over1-\tan A\tan B}}$.}

\item\label{exer:2.2.25} Solve $\dst

{y'\sqrt{1-x^2}+\sqrt{1-y^2}=0}$ explicitly.

\hint{Use the identity

$\sin(A-B)=\sin A\cos B-\cos A\sin B$.}

\item\label{exer:2.2.26}

Solve $\dst{y'={\cos x\over \sin y},\quad y (\pi)={\pi\over2}}$

explicitly. \hint{Use the identity $\cos(x+\pi/2)=-\sin x$ and the

periodicity of the cosine.}

\item\label{exer:2.2.27}

Solve the initial value problem

$$

y'=ay-by^2,\quad y(0)=y_0.

$$

Discuss the behavior of the solution if

\part{a} $y_0\ge0$; \part{b} $y_0<0$.

\item\label{exer:2.2.28}

The population $P=P(t)$ of a species satisfies

the logistic equation

$$

P'=aP(1-\alpha P)

$$

and $P(0)=P_0>0$. Find $P$ for $t>0$, and find

$\lim_{t\to\infty}P(t)$.

\item\label{exer:2.2.29}

An epidemic spreads through a population at a rate

proportional to the product of the number of people already infected

and the number of people susceptible, but not yet infected. Therefore,

if $S$ denotes the total population of susceptible people and

$I=I(t)$ denotes the number of infected people at time $t$, then

$$

I'=rI(S-I),

$$

where $r$ is a positive constant. Assuming that $I(0)=I_0$,

find $I(t)$ for $t>0$, and show that

$\lim_{t\to\infty}I(t)=S$.

\item\label{exer:2.2.30}

\Lex

The result of Exercise~\ref{exer:2.2.29} is discouraging: if any

susceptible member of the group is

initially infected, then in the long run all susceptible members are

infected!

On a more hopeful note, suppose the disease spreads

according to the model of Exercise~\ref{exer:2.2.29}, but there's a

medication that cures the infected population at a rate proportional

to the number of infected individuals. Now the equation for the number

of infected individuals becomes

$$

I'=rI(S-I)-qI

\eqno{\rm(A)}

$$

where $q$ is a positive constant.

\begin{alist}

\item % (a)

Choose $r$ and $S$ positive.

By plotting direction fields and solutions of (A) on

suitable rectangular grids

$$

R=\{0\le t \le T,\ 0\le I \le d\}

$$

in the $(t,I)$-plane, verify that if $I$ is any solution of

(A) such that $I(0)>0$, then $\lim_{t\to\infty}I(t)=S-q/r$

if $q<rS$ and $\lim_{t\to\infty}I(t)=0$ if $q\ge rS$.

\item % (b)

To verify the experimental results of \part{a}, use separation of

variables to solve (A) with initial condition $I(0)=I_0>0$, and find

$\lim_{t\to\infty}I(t)$. \hint{There are three cases to consider:

\part{i} $q<rS$; \part{ii} $q>rS$; \part{iii} $q=rS$.}

\end{alist}

\item\label{exer:2.2.31}

\Lex

Consider the differential equation

$$

y'=ay-by^2-q,

\eqno{\rm(A)}

$$

where $a$, $b$ are positive constants, and $q$ is an arbitrary

constant. Suppose $y$ denotes a solution of this equation

that satisfies the initial condition $y(0)=y_0$.

\begin{alist}

\item % (a)

Choose $a$ and $b$ positive and $q<a^2/4b$.

By plotting direction fields and solutions of (A) on

suitable rectangular grids

$$

R=\{0\le t \le T,\ c\le y \le d\}

\eqno{\rm(B)}

$$

in the $(t,y)$-plane, discover that there are numbers $y_1$ and

$y_2$ with $y_1<y_2$ such that if $y_0>y_1$ then

$\lim_{t\to\infty}y(t)=y_2$, and if $y_0<y_1$ then $y(t)=-\infty$

for some finite value of~$t$. (What happens if $y_0=y_1$?)

\item % (b)

Choose $a$ and $b$ positive and $q=a^2/4b$. By plotting direction

fields

and solutions of (A) on suitable rectangular grids of the form (B),

discover that there's a number $y_1$ such that if $y_0\ge y_1$ then

$\lim_{t\to\infty}y(t)=y_1$, while if $y_0<y_1$ then $y(t)=-\infty$

for some finite value of $t$.

\item % (c)

Choose positive $a$, $b$ and $q>a^2/4b$. By plotting direction fields

and solutions of (A) on suitable rectangular grids of the form (B),

discover that

no matter what $y_0$ is, $y(t)=-\infty$ for some finite value of $t$.

\item % (d)

Verify your results experiments analytically.

Start by separating variables in (A) to obtain

$$

{y'\over ay-by^2-q}=1.

$$

To decide what to do next you'll have to use the

quadratic formula. This should lead you to see why there are

three cases. Take it from there!

Because of its role in the transition between these

three cases, $q_0=a^2/4b$ is called a {\color{blue}\it bifurcation value\/}

of $q$. In general, if $q$ is a parameter in any differential

equation, $q_0$ is said to be a bifurcation value of

$q$ if the nature of the solutions of the equation with $q<q_0$

is qualitatively different from the nature of the solutions

with $q>q_0$.

\end{alist}

\item\label{exer:2.2.32} \Lex

By plotting direction fields and solutions of

$$

y'=qy-y^3,

$$

convince yourself that $q_0=0$ is a bifurcation value of $q$

for this equation. Explain what makes you draw this conclusion.

\item\label{exer:2.2.33}

Suppose a disease spreads according to the model of

Exercise~\ref{exer:2.2.29}, but

there's a medication that cures the infected population at a constant

rate of $q$ individuals per unit time, where $q>0$.

Then

the equation for the number of infected individuals becomes

$$

I'=rI(S-I)-q.

$$

Assuming that $I(0)=I_0>0$, use the results of

Exercise~\ref{exer:2.2.31} to describe what happens as $t\to\infty$.

\item\label{exer:2.2.34}

Assuming that $p \not\equiv 0$, state conditions under which the linear

equation

$$

y'+p(x)y=f(x)

$$

is separable. If the equation satisfies these conditions, solve it by

separation of variables and by the method developed in

Section~2.1.

\exercisetext{Solve the equations in

Exercises~\ref{exer:2.2.35}--\ref{exer:2.2.38} using variation of

parameters followed by separation of variables.}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.2.35} $\dst{y'+y={2xe^{-x}\over1+ye^x}}$&

\item\label{exer:2.2.36}\vspace*{7pt} $\dst{xy'-2y={x^6\over y+x^2}}$

\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.2.37} $\dst{y'-y}={(x+1)e^{4x}\over(y+e^x)^2}$&

\item\label{exer:2.2.38} $y'-2y=\dst{xe^{2x}\over1-ye^{-2x}}$

\end{tabular}

\item\label{exer:2.2.39}

Use variation of parameters to show that the solutions of the

following equations are of the form $y=uy_1$, where $u$ satisfies a

separable equation $u'=g(x)p(u)$. Find $y_1$ and $g$ for each

equation.

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

{\bf (a)} $xy'+y=h(x)p(xy)$

& {\bf (b)} $\dst{xy'-y=h(x) p\left({y\over x}\right)}$\\[2\jot]

{\bf (c)} $y'+y=h(x) p(e^xy)$

& {\bf (d)} $xy'+ry=h(x) p(x^ry)$\\[2\jot]

{\bf (e)} $\dst{y'+{v'(x)\over v(x)}y= h(x)

p\left(v(x)y\right)}$

\end{tabular}

\end{exerciselist}