
# 3.7: First Order Equations: Transformation of Nonlinear Equations into Separable Equations

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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

In Section~2.1 we found that the solutions of a linear
nonhomogeneous equation
$$y'+p(x)y=f(x)$$
are of the form $y=uy_1$, where $y_1$ is a nontrivial solution of the
complementary equation
\label{eq:2.4.1}
y'+p(x)y=0

and $u$ is a solution of
$$u'y_1(x)=f(x).$$
Note that this last equation is separable, since it can be rewritten
as
$$u'={f(x)\over y_1(x)}.$$
In this section we'll consider nonlinear differential equations
that are not separable to begin with, but can be solved in a similar
fashion by writing their solutions in the form $y=uy_1$, where $y_1$
is a suitably chosen known function and $u$ satisfies a separable
equation. We'llsay in this case  that we {\color{blue}\it
transformed\/} the given equation into a separable equation.

\boxit{Bernoulli Equations}

\noindent
A \href{http://www-history.mcs.st-and.ac.uk/...li_Jacob.html}
{\color{blue}\it Bernoulli
equation} is an equation  of
the form
\label{eq:2.4.2}
y'+p(x)y=f(x)y^r,

where $r$ can be any real number other than $0$ or $1$. (Note that
\eqref{eq:2.4.2} is linear if and only if $r=0$ or $r=1$.) We can
transform
\eqref{eq:2.4.2} into a separable equation by variation of parameters:
if  $y_1$ is  a nontrivial solution of \eqref{eq:2.4.1},
substituting
$y=uy_1$ into \eqref{eq:2.4.2} yields
$$u'y_1+u(y_1'+p(x)y_1)=f(x)(uy_1)^r,$$
which is equivalent to the separable equation
$$u'y_1(x)=f(x)\left(y_1(x)\right)^ru^r\mbox{\quad or \quad} {u'\over u^r}=f(x)\left(y_1(x)\right)^{r-1},$$
since $y_1'+p(x)y_1=0$.

\begin{example}\label{example:2.4.1} \rm
Solve the Bernoulli equation
\label{eq:2.4.3}
y'-y=xy^2.

\end{example}

\solution
Since $y_1=e^x$ is a solution of $y'-y=0$, we look for solutions of
\eqref{eq:2.4.3}  in the form $y=ue^x$, where
$$u'e^x=xu^2e^{2x}\mbox{\quad or, equivalently, \quad} u'=xu^2e^x.$$
Separating variables yields
$${u'\over u^2}=xe^x,$$
and integrating yields
$$-{1\over u}=(x-1)e^x+c.$$
Hence,
$$u=-{1\over(x-1)e^x+c}$$
and
$$y=-{1\over x-1+ce^{-x}}.$$

Figure~\ref{eq:2.4.1}  shows
direction field and some integral curves of \eqref{eq:2.4.3}.

\begin{figure}[tbp]
\centering
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020401}
\color{blue}
\caption{A direction field and integral curves for $y'-y=xy^{2}$}
\label{figure:2.4.1}
\end{figure}

\boxit{Other Nonlinear Equations That Can be  Transformed Into
Separable Equations}

\noindent
We've seen that the nonlinear Bernoulli equation can be transformed
into a separable equation by  the substitution $y=uy_1$ if
$y_1$ is suitably chosen. Now let's discover a sufficient condition
for a nonlinear first order differential equation
\label{eq:2.4.4}
y'=f(x,y)

to be transformable into a separable equation in the same way.
Substituting $y=uy_1$  into
\eqref{eq:2.4.4} yields
$$u'y_1(x)+uy_1'(x)=f(x,uy_1(x)),$$
which is equivalent to
\label{eq:2.4.5}
u'y_1(x)=f(x,uy_1(x))-uy_1'(x).

If
$$f(x,uy_1(x))=q(u)y_1'(x)$$
for some function $q$, then   \eqref{eq:2.4.5} becomes
\label{eq:2.4.6}
u'y_1(x)=(q(u)-u)y_1'(x),

which is separable. After checking for constant solutions $u\equiv u_0$ such that $q(u_0)=u_0$, we can separate
variables to obtain
$${u'\over q(u)-u}={y_1'(x)\over y_1(x)}.$$

\boxit{Homogeneous Nonlinear Equations}

\noindent
In the text  we'll consider only the most widely studied class of
equations for which the method of the preceding paragraph works.
Other types of equations appear in
Exercises~\ref{exer:2.4.44}--\ref{exer:2.4.51}.

The differential equation \eqref{eq:2.4.4}
is said to be {\color{blue}\it homogeneous\/} if  $x$ and $y$
occur in $f$ in such a way that  $f(x,y)$ depends only
on the ratio $y/x$; that is, \eqref{eq:2.4.4} can be written as
\label{eq:2.4.7}
y'=q(y/x),

where $q=q(u)$ is a function of a single variable.
For example,
$$y'={y+xe^{-y/x}\over x}={y\over x}+e^{-y/x}$$
and
$$y'={y^2+xy-x^2\over x^2}=\left(y\over x\right)^2+{y\over x} -1$$
are of the form  \eqref{eq:2.4.7}, with
$$q(u)=u+e^{-u}\mbox{\quad and \quad} q(u)=u^2+u-1,$$
respectively. The general method discussed above can be
applied to
\eqref{eq:2.4.7} with $y_1=x$ (and therefore $y_1'=1)$. Thus,
substituting $y=ux$ in \eqref{eq:2.4.7} yields
$$u'x+u=q(u),$$
and separation of variables (after checking for constant
solutions $u\equiv u_0$ such that $q(u_0)=u_0$) yields
$${u'\over q(u)-u}={1\over x}.$$

Before turning to examples, we point out something that you may've have
the definition of {\color{blue}\it homogeneous equation\/} given
here isn't  the same as the definition given in Section~2.1,
where we said that a linear equation of the form
$$y'+p(x)y=0$$
is homogeneous. We make no apology for this inconsistency, since we
didn't create it     historically, {\color{blue}\it homogeneous\/} has been
used in these two inconsistent ways. The one
having to do with linear equations is the most important. This
is the only section of the book where the meaning defined here will
apply.

Since $y/x$ is in general undefined if $x=0$, we'll consider
solutions of nonhomogeneous equations only on open intervals that do
not contain the point $x=0$.

\begin{example}\label{example:2.4.2}\rm
Solve
\label{eq:2.4.8}
y'={y+xe^{-y/x}\over x}.

\end{example}

\solution Substituting  $y=ux$
into \eqref{eq:2.4.8} yields
$$u'x+u = {ux+xe^{-ux/x}\over x} = u+e^{-u}.$$
Simplifying and separating variables yields
$$e^uu'={1\over x}.$$
Integrating yields
$e^u=\ln |x|+c$.
Therefore
$u=\ln(\ln|x|+c)$ and
$y=ux=x \ln (\ln |x|+c)$.

Figure~\ref{figure:2.4.2} shows
a direction field and integral curves for \eqref{eq:2.4.8}.

\begin{figure}[H]
\centering
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020402}
\color{blue}
\caption
{A direction field and some integral curves for
$y'=\dst{\frac{y+xe^{-y/x}}{x}}$}
\label{figure:2.4.2}
\end{figure}

\begin{example}\label{example:2.4.3}\rm \mbox{}\newline
\begin{alist}
\item %(a)
Solve
\label{eq:2.4.9}
x^2y'=y^2+xy-x^2.

\item %(b)
Solve the initial value problem
\label{eq:2.4.10}

\end{alist}
\end{example}

\solutionpart{a}
We first find solutions of \eqref{eq:2.4.9} on open intervals that don't
contain $x=0$. We can rewrite \eqref{eq:2.4.9} as
$$y'={y^2+xy-x^2\over x^2}$$
for $x$ in any such interval. Substituting $y=ux$ yields
$$u'x+u ={ (ux)^2+x(ux)-x^2 \over x^2} = u^2+u-1,$$
so
\label{eq:2.4.11}
u'x=u^2-1.

By inspection this equation has the constant solutions $u\equiv1$ and
$u\equiv-1$. Therefore $y=x$ and $y=-x$ are solutions of
\eqref{eq:2.4.9}. If $u$ is a solution of \eqref{eq:2.4.11} that doesn't
assume the values $\pm 1$ on some interval,  separating variables
yields
$${u'\over u^2-1}={1\over x},$$
or, after a partial fraction expansion,
$${1\over 2}\left[{1\over u-1}-{1\over u+1}\right]u'= {1\over x}.$$
Multiplying by 2 and integrating yields
$$\ln\left|u-1\over u+1\right| =2 \ln |x|+k,$$
or
$$\left|{u-1\over u+1}\right|=e^kx^2,$$
which holds if
\label{eq:2.4.12}
{u-1\over u+1}=cx^2

where $c$ is an arbitrary constant.
Solving for $u$ yields
$$u ={1+cx^2\over 1-cx^2}.$$

\begin{figure}[htbp]
\color{blue}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.6}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020403} }
\color{blue}
\caption{A direction field and  integral curves for
$x^{2}y'=y^{2}+xy-x^{2}$}
\label{figure:2.4.3}
\end{minipage}
\hspace{0.6cm}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.7}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020404}}
\caption{Solutions of  $x^{2}y'=y^{2}+xy-x^{2}$,\; $y(1)=2$}
\label{figure:2.4.4}
\end{minipage}
\end{figure}

\noindent
Therefore
\label{eq:2.4.13}
y=ux={x(1+cx^2)\over 1-cx^2}

is a solution of \eqref{eq:2.4.10} for any choice of the constant $c$.
Setting $c=0$ in \eqref{eq:2.4.13} yields the solution $y=x$. However, the
solution $y=-x$ can't be obtained from \eqref{eq:2.4.13}. Thus, the
solutions of \eqref{eq:2.4.9} on intervals that don't contain $x=0$ are
$y=-x$ and functions of the form \eqref{eq:2.4.13}.

The situation is more complicated if $x=0$ is the open interval.
First, note that $y=-x$ satisfies \eqref{eq:2.4.9}
on $(-\infty,\infty)$. If $c_1$ and $c_2$ are arbitrary constants,
the function
\label{eq:2.4.14}
y=\left\{\begin{array}{ll} \dst{x(1+c_1x^2)\over 1-c_1x^2},&a<x<0,\\
[2\jot]\dst{x(1+c_2x^2)\over 1-c_2x^2},&0\le x<b,
\end{array}\right.

is a solution of \eqref{eq:2.4.9} on $(a,b)$, where
$$a=\left\{\begin{array}{cl}-\dst{1\over\sqrt{c_1}}&\mbox{ if }c_1>0,\\ -\infty&\mbox{ if }c_1\le 0, \end{array}\right. \mbox{\quad and \quad} b=\left\{\begin{array}{cl}\dst{1\over\sqrt{c_2}}&\mbox{ if }c_2>0,\\ \infty&\mbox{ if }c_2\le 0. \end{array}\right.$$
We leave it to you to verify this. To do so, note that if $y$ is
any function of the form \eqref{eq:2.4.13} then $y(0)=0$ and $y'(0)=1$.

Figure~\ref{figure:2.4.3} shows a direction field and some integral curves
for \eqref{eq:2.4.9}.

\solutionpart{b}  We could obtain $c$ by imposing
the initial condition $y(1)=2$ in \eqref{eq:2.4.13}, and then solving for
$c$. However, it's easier to use \eqref{eq:2.4.12}. Since $u=y/x$, the
initial
condition
$y(1)=2$ implies that $u(1)=2$.  Substituting this into \eqref{eq:2.4.12}
yields $c=1/3$.  Hence, the solution of \eqref{eq:2.4.10} is
$$y={x(1+x^2/3)\over 1-x^2/3}.$$
The interval of validity of this solution is $(-\sqrt3,\sqrt3)$.
However, the largest interval on which \eqref{eq:2.4.10} has a unique
solution is $(0,\sqrt3)$. To see this, note from \eqref{eq:2.4.14}
that any function of the form
\label{eq:2.4.15}
y=\left\{\begin{array}{ll} \dst{x(1+cx^2)\over
1-cx^2},&a<x\le0,\\[2\jot]
\dst{x(1+x^2/3)\over 1-x^2/3},&0\le x<\sqrt3,
\end{array}\right.

is a solution of \eqref{eq:2.4.10} on $(a,\sqrt3)$, where $a=-1/\sqrt c$
if $c>0$ or $a=-\infty$ if $c\le0$. (Why doesn't this contradict
Theorem~\ref{thmtype:2.3.1}?)

Figure~\ref{figure:2.4.4} shows several solutions of the
initial value problem~\eqref{eq:2.4.10}. Note that these solutions coincide
on $(0,\sqrt{3})$.

In the last two examples we were able to solve the given equations
explicitly.   However, this isn't  always possible, as you'll
see in the exercises.

\exercises
In Exercises~\ref{exer:2.4.1}--\ref{exer:2.4.4} solve the given Bernoulli
equation.

\begin{exerciselist}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.1} $y'+y=y^2$ &
\item\label{exer:2.4.2} $\dst {7xy'-2y=-{x^2 \over y^6}}$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.3} $x^2y'+2y=2e^{1/x}y^{1/2}$
& \item\label{exer:2.4.4} \vspace*{-5pt}$\dst {(1+x^2)y'+2xy ={1 \over (1+x^2)y}}$
\end{tabular}

\exercisetext{In Exercises~\ref{exer:2.4.5} and \ref{exer:2.4.6} find all
solutions. Also, plot a direction field and some integral curves on
the indicated rectangular region.}

\item\label{exer:2.4.5}  \CGex
$y'-xy=x^3y^3; \quad \{-3\le x\le 3,\-2\le y\ge 2\}$

\item\label{exer:2.4.6} \CGex
$\dst {y'-{1+x\over 3x}y=y^4}; \quad \{-2\le x\le2,-2\le y \le2\}$

\exercisetext{In Exercises~\ref{exer:2.4.7}--\ref{exer:2.4.11} solve the
initial value problem.}

\item\label{exer:2.4.7} $y'-2y=xy^3,\quad y(0)=2\sqrt2$

\item\label{exer:2.4.8} $y'-xy=xy^{3/2},\quad y(1)=4$
\item\label{exer:2.4.9} $xy'+y=x^4y^4,\quad y(1)=1/2$
\item\label{exer:2.4.10} $y'-2y=2y^{1/2},\quad y(0)=1$
\item\label{exer:2.4.11} $\dst{y'-4y={48x\over y^2},\quad y(0)=1}$

\exercisetext{In Exercises~\ref{exer:2.4.12} and \ref{exer:2.4.13} solve the
initial value problem and graph the solution.}

\item\label{exer:2.4.12} \CGex $x^2y'+2xy=y^3,\quad y(1)=1/\sqrt2$
\item\label{exer:2.4.13} \CGex $y'-y=xy^{1/2},\quad y(0)=4$

\item\label{exer:2.4.14}
You may have noticed that  the logistic equation
$$P'=aP(1-\alpha P)$$
from Verhulst's model for population growth can be written in
Bernoulli form as
$$P'-aP=-a\alpha P^2.$$
This isn't particularly interesting, since the logistic equation is
separable, and therefore solvable by the method studied in
Section~2.2. So let's consider a more complicated model,
where
$a$ is a positive constant and $\alpha$ is a positive continuous
function of $t$ on $[0,\infty)$. The equation for this model is
$$P'-aP=-a\alpha(t) P^2,$$
a non-separable Bernoulli equation.
\begin{alist}
\item % (a)
Assuming that $P(0)=P_0>0$, find $P$ for $t>0$. \hint{Express your
result in terms of the integral
$\int_0^t\alpha(\tau)e^{a\tau}\,d\tau$.}
\item % (b)
Verify that your result reduces to the known results
for the Malthusian model where  $\alpha=0$, and the Verhulst model
where $\alpha$ is a nonzero constant.
\item % (c)
Assuming that
$$\lim_{t\to\infty}e^{-at}\int_0^t\alpha(\tau)e^{a\tau}\,d\tau=L$$
exists (finite or infinite), find $\lim_{t\to\infty}P(t)$.
\end{alist}

\exercisetext{In Exercises~\ref{exer:2.4.15}--\ref{exer:2.4.18} solve
the equation explicitly.}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.15} $y'=\dst{y+x\over x}$ & \vspace*{5pt} \item\label{exer:2.4.16}
$y'=\dst{y^2+2xy \over x^2}$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.17} $xy^3y'=y^4+x^4$ &
\item\label{exer:2.4.18} $y'=\dst{y\over x}+\sec{y\over x}$
\end{tabular}

\exercisetext{In Exercises~\ref{exer:2.4.19}-\ref{exer:2.4.21} solve the
equation explicitly.
Also, plot a direction field and some integral curves on
the indicated rectangular region.}

\item\label{exer:2.4.19} \CGex
$x^2y'=xy+x^2+y^2; \quad \{-8\le x\le 8,-8\le y\le 8\}$

\item\label{exer:2.4.20} \CGex
$xyy'=x^2+2y^2; \quad \{-4\le x\le 4,-4\le y\le 4\}$

\item\label{exer:2.4.21} \CGex
$y'=\dst{2y^2+x^2e^{-(y/x)^2}\over 2xy}; \quad \{-8\le x\le 8,-8\le y\le 8\}$

\exercisetext{In Exercises~\ref{exer:2.4.22}--\ref{exer:2.4.27} solve
the initial value problem.}

\item\label{exer:2.4.22} $y'=\dst{xy+y^2\over x^2}, \quad y(-1)=2$

\item\label{exer:2.4.23} $y'=\dst{x^3+y^3\over xy^2}, \quad y(1)=3$

\item\label{exer:2.4.24}
$xyy'+x^2+y^2=0, \quad y(1)=2$

\item\label{exer:2.4.25}
$y'=\dst{y^2-3xy-5x^2 \over x^2}, \quad y(1)=-1$

\item\label{exer:2.4.26}
$x^2y'=2x^2+y^2+4xy, \quad y(1)=1$

\item\label{exer:2.4.27}
$xyy'=3x^2+4y^2, \quad y(1)=\sqrt{3}$

\exercisetext{In Exercises~\ref{exer:2.4.28}--\ref{exer:2.4.34} solve the
given homogeneous equation implicitly.}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.28} $y'=\dst{x+y \over x-y}$&
\item\label{exer:2.4.29}\vspace*{13pt} $(y'x-y)(\ln |y|-\ln |x|)=x$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.30} $y'=\dst{y^3+2xy^2+x^2y+x^3\over x(y+x)^2}$ &
\item\label{exer:2.4.31}  $y'=\dst{x+2y \over 2x+y}$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.32}  $y'=\dst{y \over y-2x}$&
\item\label{exer:2.4.33} $y'=\dst{xy^2+2y^3\over x^3+x^2y+xy^2}$
\end{tabular}

\item\label{exer:2.4.34} $y'=\dst{x^3+x^2y+3y^3 \over x^3+3xy^2}$

\item\label{exer:2.4.35} \Lex
\begin{alist}
\item % (a)
Find a solution of the initial value problem
$$x^2y'=y^2+xy-4x^2, \quad y(-1)=0 \eqno{\rm(A)}$$
on the interval $(-\infty,0)$. Verify that this solution is actually
valid on $(-\infty,\infty)$.
\item % (b)
Use Theorem~\ref{thmtype:2.3.1} to show that (A) has a
unique solution on $(-\infty,0)$.
\item % (c)
Plot a direction field for the differential equation in
(A) on a square
$$\{-r\le x\le r, -r\le y\le r\},$$
where $r$
is any positive number. Graph the solution you obtained in
\part{a} on this field.
\item % (d)
Graph  other solutions  of (A) that are defined on $(-\infty,\infty)$.
\item % (e)
Graph other solutions of (A)  that are defined only on intervals of
the form $(-\infty,a)$, where is a finite positive number.
\end{alist}

\item\label{exer:2.4.36} \Lex
\begin{alist}
\item % (a)
Solve the equation
$$xyy'=x^2-xy+y^2 \eqno{\rm(A)}$$
implicitly.
\item % (b)
Plot a direction field for (A) on a square
$$\{0\le x\le r,0\le y\le r\}$$
where $r$ is any positive number.
\item % (c)
Let $K$ be a positive integer. (You may have to try several
choices for $K$.)
Graph solutions of the initial value problems
$$xyy'=x^2-xy+y^2,\quad y(r/2)={kr\over K},$$
for $k=1$, $2$, \dots, $K$. Based on your observations, find
conditions on the positive numbers $x_0$
and $y_0$ such that the initial value problem
$$xyy'=x^2-xy+y^2,\quad y(x_0)=y_0, \eqno{\rm(B)}$$
has  a unique solution (i) on $(0,\infty)$   or (ii) only
on an interval  $(a,\infty)$, where $a>0$?
\item % (d)
What can you say about the graph of the solution of (B)
as $x\to\infty$? (Again, assume that $x_0>0$ and $y_0>0$.)
\end{alist}

\item\label{exer:2.4.37} \Lex
\begin{alist}
\item % (a)
Solve the equation
$$y'={2y^2-xy+2x^2 \over xy+2x^2} \eqno{\rm(A)}$$
implicitly.
\item % (b)
Plot a direction field for (A) on a square
$$\{-r\le x\le r,-r\le y\le r\}$$
where $r$ is any positive number. By graphing solutions of (A),
determine necessary and sufficient conditions on $(x_0,y_0)$ such that
(A) has a solution on (i) $(-\infty,0)$ or (ii) $(0,\infty)$ such that
$y(x_0)=y_0$.
\end{alist}

\item\label{exer:2.4.38} \Lex
Follow the instructions of Exercise~\ref{exer:2.4.37} for the equation
$$y'={xy+x^2+y^2 \over xy}.$$

\item\label{exer:2.4.39} \Lex
Pick any  nonlinear homogeneous equation $y'=q(y/x)$  you like, and
plot  direction fields on the square $\{-r\le x\le r,\ -r\le y\le r\}$,
where $r>0$. What happens to the direction field as you vary $r$?
Why?

\item\label{exer:2.4.40}
Prove:  If $ad-bc\ne 0$, the equation
$$y'={ax+by+\alpha \over cx+dy+\beta}$$
can be transformed into the homogeneous nonlinear equation
$${dY \over dX}={aX+bY \over cX+dY}$$
by the substitution $x=X-X_0,\ y=Y-Y_0$,
where $X_0$ and $Y_0$ are suitably chosen constants.

\exercisetext{In Exercises~\ref{exer:2.4.41}-\ref{exer:2.4.43} use a
method suggested by Exercise~\ref{exer:2.4.40} to solve the
given equation implicitly.}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.41} $y'=\dst{-6x+y-3 \over 2x-y-1}$
& \item\label{exer:2.4.42}\vspace*{10pt} $y'=\dst{2x+y+1 \over x+2y-4}$
\end{tabular}

\item\label{exer:2.4.43}
$y'=\dst{-x+3y-14 \over x+y-2}$

\exercisetext{In Exercises~\ref{exer:2.4.44}--\ref{exer:2.4.51} find
a function $y_1$ such that the substitution $y=uy_1$ transforms
the given equation  into a separable
equation of the form \eqref{eq:2.4.6}. Then solve the given equation
explicitly.}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.44} $3xy^2y'=y^3+x$ & \vspace*{10pt}
\item\label{exer:2.4.45} $xyy'=3x^6+6y^2$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.46} $x^3y'=2(y^2+x^2y-x^4)$ &
\item\label{exer:2.4.47} $y'=y^2e^{-x}+4y+2e^x$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.48} $y'=\dst{y^2+y\tan x+\tan^2 x\over\sin^2x}$ &
\item\label{exer:2.4.49} $x(\ln x)^2y'=-4(\ln x)^2+y\ln x+y^2$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:2.4.50} $2x(y+2\sqrt x)y'=(y+\sqrt x)^2$&
\item\label{exer:2.4.51} $(y+e^{x^2})y'=2x(y^2+ye^{x^2}+e^{2x^2})$
\end{tabular}

\medskip
\item\label{exer:2.4.52}
Solve the initial value problem
$$y'+{2\over x}y={3x^2y^2+6xy+2\over x^2(2xy+3)},\quad y(2)=2.$$

\item\label{exer:2.4.53}
Solve the initial value problem
$$y'+{3\over x}y={3x^4y^2+10x^2y+6\over x^3(2x^2y+5)},\quad y(1)=1.$$

\item\label{exer:2.4.54}
Prove:  If $y$ is a solution of a homogeneous nonlinear equation
$y'=q(y/x)$,  so is $y_1=y(ax)/a$, where $a$ is any nonzero
constant.

\item\label{exer:2.4.55}
A {\color{blue}\it generalized}
\href{http://http://www-history.mcs.st-and.ac.uk/Indexes/Riccati.html}
{\color{blue}\it Riccati equation} is of the form
$$y'=P(x)+Q(x)y+R(x)y^2. \eqno{\rm (A)}$$
(If $R\equiv-1$,  (A) is a
\href{http://http://www-history.mcs.st-and.ac.uk/Indexes/Riccati.html}
{\color{blue}\it Riccati
equation\/}.) Let $y_1$ be a known solution and $y$ an arbitrary
solution of (A). Let $z=y-y_1$. Show that $z$ is a
solution of a Bernoulli equation with $n=2$.

\exercisetext{In Exercises~\ref{exer:2.4.56}--\ref{exer:2.4.59},
given that $y_1$ is a solution of the given equation, use the
method suggested by Exercise \ref{exer:2.4.55} to find other solutions.}

\item\label{exer:2.4.56} $y'=1+x - (1+2x)y+xy^2$;  \quad    $y_1=1$

\item\label{exer:2.4.57} $y'=e^{2x}+(1-2e^x)y+y^2$;  \quad    $y_1=e^x$

\item\label{exer:2.4.58} $xy'=2-x+(2x-2)y-xy^2$;  \quad    $y_1=1$

\item\label{exer:2.4.59} $xy'=x^3+(1-2x^2)y+xy^2$;  \quad    $y_1=x$

\end{exerciselist}