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%TCIDATA{<META NAME="Title" CONTENT="Differential Equations">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 16:43:42 1997}
%TCIDATA{Language=American English}
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\begin{document}


\section{\label{Differential equation}Differential Equations}

A \emph{differential equation} is an equation that includes differentials or
derivatives. A \emph{solution} to a differential equation is any function
that satisfies the given equation. Thus $f(x)=\sin x$ is a solution to the
differential equation $y^{\prime \prime }+y=0$, because if $y=\sin x$, then $%
y^{\prime }=\cos x$ and $y^{\prime \prime }=-\sin x$, and hence $y^{\prime
\prime }+y=\sin x-\sin x=0$. Differential equations are encountered in the
study of problems in both pure and applied mathematics, in the sciences, in
engineering, and in business and the social sciences.\medskip

With the choices on the \textsf{Solve ODE} submenu you will be able to find
closed-form solutions to many differential equations. The solution is
generally returned as an equation in $y(x)$ and $x$ (or whatever variables
were specified) with any arbitrary constants represented as $C_{1},$ $C_{2}$%
, ..., $C_{n}$\medskip $.$

\begin{quote}
$\blacktriangleright $ To solve a differential equation 
\index{Solve@Solve!differential equation@differential equation}
\end{quote}

\begin{enumerate}
\item[1.]  Type the differential equation using standard mathematical
notation.

\item[2.]  With the insertion point in the equation, from the \textsf{Solve
ODE} submenu, choose \textsf{Exact}, \textsf{Laplace}, \textsf{Numeric}, or 
\textsf{Series.}
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Solve ODE + Exact}
\end{quote}

\begin{quotation}
$%
\dfrac{dy}{dx}=xy$, Exact solution is : $y\left( x\right) =e^{\frac{1}{2}%
x^{2}}C_{1}$
\end{quotation}

To check this result, define $y\left( x\right) =e^{\frac{1}{2}x^{2}}C_{1}$.
Replace $y$ by $y(x)$ in the differential equation and evaluate both sides.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\dfrac{dy\left( x\right) }{dx}=\allowbreak xe^{\frac{1}{2}x^{2}}C_{1}$

$xy\left( x\right) =\allowbreak xe^{\frac{1}{2}x^{2}}C_{1}\medskip $
\end{quotation}

For any given number\ $C_{1}$, the solution describes a curve. Since $C_{1}$
may, in general, take on infinitely many values there are an infinity of
solution curves, or a \emph{one-parameter family\label{One-parameter family}}
of solution curves for this equation.$\bigskip $

\subsection{\ Exact Solutions\label{Solve ODE}\label{Solve ODE, exact}}

Two methods, \textsf{Exact} and \textsf{Laplace}, return exact solutions to
a differential equation. \textsf{Laplace}, which as its name suggests, uses
the Laplace transform to derive solutions, works for either homogeneous 
\index{Differential equations@Differential equations!homogeneous@homogeneous}%
or nonhomogeneous linear 
\index{Differential Equation@Differential equations!linear@linear}%
differential equations with constant \label{Differential equations, linear}%
coefficients. \label{Differential equations, Laplace}\label%
{Differential equations, homogeneous}Initial conditions are displayed in the
solution. \textsf{Exact} is a more general method that works for some
nonlinear differential equations as well. Each of these options recognizes
some functions that the other may not.

\subsubsection{Exact Method}

When a prime indicates differentiation, the independent variable will be
named if it is unambiguous; otherwise, a variable name must be specified. In
the equations $y^{\prime }=y$, $y^{\prime }=\sin x$ and $y^{\prime }=\sin
x+t $, the independent variable is ambiguous and a dialog box appears asking
for the independent variable. When a notation is used for differentiation
that names the independent variable, it is taken from context and no dialog
box appears. 
\index{Solve ODE@Solve ODE!exact@exact}\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Solve ODE + Exact}
\end{quote}

\begin{quotation}
$y^{\prime }=\sin x$ (Specify $x$), Exact solution is: $y\left( x\right)
=-\cos x+C_{1}$

$y^{\prime }=\sin x$ (Specify $t$), Exact solution is: $y\left( t\right)
=\left( \sin x\right) t+C_{1}$

$y^{\prime }=\sin x+t$ (Specify $x$), Exact solution is: $y\left( x\right)
=-\cos x+tx+C_{1}$

$y^{\prime }=\sin x+t$ (Specify $t$), Exact solution is: $y\left( t\right)
=\left( \sin x\right) t+%
\frac{1}{2}t^{2}+C_{1}$

$y^{\prime }=y$, Exact solution is: $y\left( t\right) =e^{t}C_{1}\medskip $
\end{quotation}

The constant $C_{1}$ indicates that there is a family of solutions, one for
each choice of $C_{1}$. The following figure shows solutions for $y^{\prime
}=y$ corresponding to the choices $1/2$, $1$, $2$, $3$, and $4$ for $C_{1}$.
To replicate this plot, drag solutions to the frame one at a time and then
edit the colors on the \textsf{Plot Components} page of the \textsf{Plot
Properties} dialog.

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"normal";xis \TEXUX{t};var1name \TEXUX{$x$};}}

\textsl{Scientific Notebook} recognizes a variety of notation for a
differential equation. The following examples illustrate some of this
variety. The Leibniz notation $\frac{dy}{dx}$ or the $D_{x}$ notation
provides \textsl{Scientific Notebook} with enough information to determine
the independent variable without a dialog box.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Solve ODE + Exact}
\end{quote}

\begin{quotation}
$\dfrac{dy}{dx}=y+x$, Exact solution is: $y\left( x\right) =-x-1+e^{x}C_{1}$

$y^{\prime }+xy=x$ (Specify $x$), Exact solution is: $y\left( x\right)
=1+e^{-\frac{1}{2}x^{2}}C$\medskip

$D_{x}y-y=\sin x$, Exact solution is: $y\left( x\right) =-\frac{1}{2}\cos x-%
\frac{1}{2}\sin x+e^{x}C_{1}$\medskip
\end{quotation}

Following is a plot of three particular solutions for $D_{x}y-y=\sin x$
corresponding to $C_{1}=1,2,3$. To replicate this plot, drag solutions to
the frame one at a time and then edit the colors on the \textsf{Plot
Components} page of the \textsf{Plot Properties} dialog.$\medskip $

\begin{quote}
$\blacktriangleright $ \textsf{Plot 2D + Rectangular}
\end{quote}

\begin{quotation}
$-\frac{1}{2}\cos x-\frac{1}{2}\sin x+e^{x},-\frac{1}{2}\cos x-\frac{1}{2}%
\sin x+2e^{x},-\frac{1}{2}\cos x-\frac{1}{2}\sin x+3e^{x}$\FRAME{dtbpF}{3in}{%
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\TEXUX{$-\frac{1}{2}\cos x-\frac{1}{2}\sin x+2e^{x}$};linecolor
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49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};}}
\end{quotation}

The three solutions can be distinguished by evaluation at 0. For example,
the solution with $C_1=1$ crosses the $y$-axis at $y=1/2$.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Solve ODE + Exact}
\end{quote}

\begin{quotation}
$y^{\prime \prime }+y=x^{2}$ (Specify $x$), Exact solution is: $y\left(
x\right) =-2+x^{2}+C_{1}\cos x+C_{2}\sin x\medskip $
\end{quotation}

The following plot shows three solutions generated with constants $\left(
C_{1},C_{2}\right) =\left( 1,1\right) $, $\left( C_{1},C_{2}\right) =\left(
5,1\right) $, and $\left( C_{1},C_{2}\right) =\left( 1,5\right) $. To
replicate this plot, drag solutions to the frame one at a time and then edit
the colors on the \textsf{Plot Components} page of the \textsf{Plot
Properties} dialog.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Plot 2D + Rectangular}
\end{quote}

\begin{quotation}
$x^{2}-2+\sin x+\cos x,x^{2}-2+5\sin x+\cos x,x^{2}-2+\sin x+5\cos x$
\end{quotation}

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\TEXUX{x};var1name \TEXUX{$x$};}}

The particular solution $y(x)=x^2-2+\sin x+5\cos x$ is the one whose graph
crosses the $y$-axis at $y=3$.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Solve ODE + Exact}
\end{quote}

\begin{quotation}
$\dfrac{d^{2}y}{dx^{2}}+\dfrac{dy}{dx}=x+y$, Exact solution is: $y\left(
x\right) =-1-x+C_{1}e^{\frac{1}{2}\left( -1+\sqrt{5}\right) x}+C_{2}e^{-%
\frac{1}{2}\left( 1+\sqrt{5}\right) x}\medskip $

$xy^{\prime }-y=x^{2}$, Exact solution is : $y\left( x\right)
=x^{2}+xC_{1}\bigskip $
\end{quotation}

Some differential equations that are not readily solvable by this method can
be solved after rewriting the equation, as shown in the following two
examples.\medskip

\begin{example}
The differential equation $\frac{dy}{dx}=\frac{1}{x^{2}\sin y-xy}$ can be
solved by reversing the role of the two variables.

\begin{itemize}
\item  \textsf{Solve ODE + Exact }$\frac{dx}{dy}=x^{2}\sin y-xy$, Exact
solution is : $\frac{1}{x\left( y\right) }=\left( -\int e^{-\frac{1}{2}%
y^{2}}\sin y\,dy+C_{1}\right) e^{\frac{1}{2}y^{2}}$
\end{itemize}
\end{example}

\begin{example}
The differential equation $\frac{dy}{dx}+\frac{xy}{1-x^{2}}=x\sqrt{y}\ $can
be solved after a change of variable. Take $\sqrt{y}=z$ to get $2z\frac{dz}{%
dx}+\frac{xz^{2}}{1-x^{2}}=xz$.

\begin{itemize}
\item  \textsf{Solve ODE + Exact}, \textsf{Simplify }$\frac{dz}{dx}+\frac{xz%
}{2\left( 1-x^{2}\right) }=\frac{x}{2}$, Exact solution is : $z\left(
x\right) =\frac{1}{3}x^{2}-\frac{1}{3}+\sqrt[4]{\left( 2x^{2}-2\right) }%
C_{1} $
\end{itemize}

Thus $y(x)=\left( \frac{1}{3}x^{2}-\frac{1}{3}+\sqrt[4]{\left(
2x^{2}-2\right) }C_{1}\right) ^{2}$. To check this result, define $y(x)$ and
evaluate the two sides of the differential equation to get $\frac{d}{dx}%
\left( y\left( x\right) \right) +\frac{xy\left( x\right) }{1-x^{2}}%
=\allowbreak \frac{1}{3}\left( -1+x^{2}\right) x$ and $x\sqrt{y\left(
x\right) }=\allowbreak \frac{1}{3}\left( -1+x^{2}\right) x$.
\end{example}

\subsubsection{Laplace Method 
\index{Solve ODE@Solve ODE!Laplace@Laplace}\label{Solve ODE, Laplace}\label%
{Laplace}}

Laplace transforms solve either homogeneous 
\index{Differential equations@Differential equations!homogeneous@homogeneous}%
or nonhomogeneous \label{Nonhomogeneous linear ODE}linear\label%
{Homogeneous ODE} systems in which the coefficients are all constants.
Initial conditions appear explicitly in the solution.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Solve ODE + Laplace}
\end{quote}

\begin{quotation}
$%
\dfrac{dy}{dx}=y$, Laplace solution is: $y\left( x\right) =y\left( 0\right)
e^{x}$

$\mathsf{\,\,}y^{\prime }+y=x+\sin x$ (Specify $x$), Laplace solution is:

\qquad \qquad \qquad $y\left( x\right) =-1+x+\frac{3}{2}e^{-x}+y\left(
0\right) e^{-x}+\frac{1}{2}\sin x-\frac{1}{2}\cos x\medskip $
\end{quotation}

The following examples compare exact and Laplace solutions.

\begin{center}
\begin{tabular}{lll}
\textbf{Equation} & \textbf{Exact} & \textbf{Laplace} \\ \hline
$y^{\prime }=\sin x$ & $y\left( x\right) =-\cos x+C_{1}$ & $y\left( x\right)
=1+y\left( 0\right) -\cos x$ \\ 
$y^{\prime }=y+x$ & $y\left( x\right) =-x-1+e^{x}C_{1}$ & $y\left( x\right)
=-1-x+e^{x}+y\left( 0\right) e^{x}$ \\ 
$D_{x}y=\cos x$ & $y\left( x\right) =\sin x+C_{1}$ & $y\left( x\right)
=y\left( 0\right) +\sin x$ \\ 
$D_{x}y=x+t$ & $y\left( x\right) =\frac{1}{2}x^{2}+tx+C_{1}$ & $y\left(
x\right) ={\frac{1}{2}}x^{2}+y\left( 0\right) +tx$ \\ 
$\frac{dy}{dx}=y$ & $y\left( x\right) =e^{x}C_{1}$ & $y\left( x\right)
=y\left( 0\right) e^{x}$ \\ 
$y^{\prime }=y^{2}+1$ & $-\arctan \left( y\left( t\right) \right) +t=C_{1}$
& $\text{Fails}$ \\ 
$\left( y^{\prime }\right) ^{3}-3\left( y^{\prime }\right) ^{2}+2y^{\prime
}=0$ & $y\left( t\right) =C_{1},\allowbreak y\left( t\right)
=2t+C_{1},\allowbreak y\left( t\right) =t+C_{1}$ & $\text{Fails}$%
\end{tabular}
\end{center}

\strut

\subsection{\ Series Solutions\label{Series solution to ODE}\label%
{Settings, series order}\label{Solve ODE, series}}

For many applications, a few terms of a Taylor \label{ODE, series solution}%
series 
\index{Solve ODE@Solve ODE!series@series} 
\index{Taylor series@Taylor series}solution are sufficient. 
\index{Series@Series!Taylor@Taylor}You can control the number of terms that
appear in the solution by changing \textsf{Series Order}\label%
{Series, Taylor}\label{Taylor series, (2)}\textsf{\ for ODE Solutions} in
the \textsf{Maple Settings} 
\index{Settings@Settings!series order@series order}dialog. For most of the
following examples, the series order has been set at 6.\medskip

\begin{quote}
$\blacktriangleright $ To specify the order of terms in a series solution
\end{quote}

\begin{enumerate}
\item  Choose \textsf{Maple + Settings, General} page.

\item  Set \textsf{Series Order for ODE Solutions} to desired order.

\item  Choose \textsf{OK.\medskip }
\end{enumerate}

In the following examples, notice that the initial condition $y(0)$ appears
explicitly in each solution. Choose \textsf{Series} from the \textsf{Solve
ODE} submenu to produce the following solutions.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Solve ODE + Series}
\end{quote}

\begin{quotation}
$D_{x}y=y$, Series solution is : $y\left( x\right) =y\left( 0\right)
+y\left( 0\right) x+\left( 
\frac{1}{2}y\left( 0\right) \right) x^{2}+\left( \frac{1}{6}y\left( 0\right)
\right) x^{3}+\left( \frac{1}{24}y\left( 0\right) \right) x^{4}+\left( \frac{%
1}{120}y\left( 0\right) \right) x^{5}+O\left( x^{6}\right) $\medskip

$y^{\prime }=\dfrac{\sin x}{x}$ (Specify $x$), Series solution is : $y\left(
x\right) =y\left( 0\right) +x-\frac{1}{18}x^{3}+\frac{1}{600}x^{5}+O\left(
x^{6}\right) \medskip $

$\dfrac{dy}{dx}+y=e^{-x^{2}}$, Series solution is : $y\left( x\right)
=y\left( 0\right) \allowbreak +\left( -y\left( 0\right) +1\right)
x\allowbreak +\left( \frac{1}{2}y\left( 0\right) -\frac{1}{2}\right)
x^{2}\allowbreak +\left( -\frac{1}{6}y\left( 0\right) -\frac{1}{6}\right)
x^{3}\allowbreak +\left( \frac{1}{24}y\left( 0\right) +\frac{1}{24}\right)
x^{4}\allowbreak +\left( -\frac{1}{120}y\left( 0\right) +\frac{11}{120}%
\right) x^{5}\allowbreak +O\left( x^{6}\right) $
\end{quotation}

\strut

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Systems of Ordinary Differential Equations}{}{}{%
DM10-2.tex#Exact solutions}

\item  \hyperref{Numerical Methods for ODEs}{}{}{DM10-3.tex#Solve ODE,
numeric}

\item  \hyperref{Partial Differential Equations}{}{}{DM10-4.tex#PDE}

\item  \hyperref{Exercises and Solutions}{}{}{DM10-5.tex#Exercises}
\end{itemize}

\rule{0.67in}{0.01in}

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"1";cropbottom "0";filename 'index.wmf';file-properties "XNPEU";}}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Differential
equations}{}{}{DMIndex.tex#Differential equations}, \hyperref{Maple settings%
}{}{}{DMIndex.tex#Maple settings}, \hyperref{Solve ODE}{}{}{%
DMIndex.tex#Solve ODE}, \hyperref{Taylor series}{}{}{DMIndex.tex#Taylor
series}}

$\;$

\end{document}
