planet/北京师范大学-香港浸会大学联合国际学院/常微分方程/Chapter_2.tm

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<\body>
  <screens|<\shown>
    <doc-data|<doc-title|Ordinary Differential
    Equations>|<doc-author|<author-data|<author-name|Yuliang Wang>>>>

    <chapter*|Chapter 2: First Order Differential Equations>

    General form\ 

    <\equation*>
      F<around*|(|u,u<rprime|'>|)>=0
    </equation*>

    <\example>
      <dueto|falling object in air, general solution, integral curves,
      initial value problem>

      The motion of the object is governed by the Newton's law. Let
      <math|v<around*|(|t|)>> be the velocity of the object at time <math|t>.

      <\equation*>
        m*v<rprime|'>=m*g-\<gamma\>*v,
      </equation*>

      where <math|m> is the mass of the object, <math|g> is the gravitational
      constant, and <math|\<gamma\>> is the coefficient of air resistant
      force.

      This is a first order linear ODE.
    </example>

    <section|Method of integrating factors>

    <\example>
      Find the general solutions of

      <\equation*>
        u<rprime|'>+u=2
      </equation*>

      idea: combine <math|u<rprime|'>+u> into the derivative of another
      function.\ 

      Consider the product rule of differentiation:

      <\equation*>
        <around*|(|f*g|)><rprime|'>=f<rprime|'>*g+f*g<rprime|'>.
      </equation*>

      Let <math|f=u<around*|(|t|)>,g=e<rsup|t>>, then

      <\equation*>
        <around*|[|u<around*|(|t|)>*e<rsup|t>|]><rprime|'>=u<rprime|'>*e<rsup|t>+u*e<rsup|t>=e<rsup|t>*<around*|(|u<rprime|'>+u|)>.
      </equation*>

      Now, we multiply the original equation by <math|e<rsup|t>>:

      <\equation*>
        <around*|(|u<rprime|'>+u|)>*e<rsup|t>=2*e<rsup|t>\<Longrightarrow\><around*|[|u<around*|(|t|)>*e<rsup|t>|]><rprime|'>=2*e<rsup|t>\<Longrightarrow\>u<around*|(|t|)>*e<rsup|t>=<big|int>2*e<rsup|t>*\<mathd\>t=2*e<rsup|t>+c
      </equation*>
    </example>

    Divide by <math|e<rsup|t>>:

    <\equation*>
      u<around*|(|t|)>=e<rsup|-t>*<around*|[|2*e<rsup|t>+c|]>=2+c*e<rsup|-t>.
    </equation*>

    <strong|Method of integrating factors:>

    Consider the 1st order linear ODE (standard form):

    <\equation*>
      <with|color|dark green|u<rprime|'><around*|(|t|)>+p<around*|(|t|)>*u<around*|(|t|)>=q<around*|(|t|)>.>
    </equation*>

    Multiply left side by <math|\<mu\><around*|(|t|)>>:

    <\equation*>
      <around*|[|u<rprime|'><around*|(|t|)>+p<around*|(|t|)>*u<around*|(|t|)>|]>*\<mu\><around*|(|t|)>=u<rprime|'>*\<mu\>+p*u*\<mu\>.
    </equation*>

    We want

    <\equation*>
      \<mu\>=g,p*\<mu\>=g<rprime|'>
    </equation*>

    i.e.

    <\equation*>
      \<mu\><rprime|'><around*|(|t|)>=p*<around*|(|t|)>*\<mu\><around*|(|t|)>\<Longrightarrow\><with|color|red|\<mu\><around*|(|t|)>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t>>.
    </equation*>

    Check (by the chain rule and fundamental theorem of calculus)

    <\equation*>
      \<mu\><rprime|'>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t>*p<around*|(|t|)>=\<mu\><around*|(|t|)>*p<around*|(|t|)>.
    </equation*>

    Derivation directly:

    <\equation*>
      \<mu\><rprime|'><around*|(|t|)>=p*<around*|(|t|)>*\<mu\><around*|(|t|)>\<Longrightarrow\><frac|\<mu\><rprime|'><around*|(|t|)>|\<mu\><around*|(|t|)>>=p<around*|(|t|)>\<Longrightarrow\><around*|[|ln*\<mu\><around*|(|t|)>|]><rprime|'>=p<around*|(|t|)>
    </equation*>

    <\equation*>
      \<Longrightarrow\>ln*\<mu\><around*|(|t|)>=<big|int>p<around*|(|t|)>*\<mathd\>t\<Longrightarrow\>\<mu\><around*|(|t|)>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t.>
    </equation*>

    So the original ODE becomes

    <\equation*>
      <around*|[|\<mu\><around*|(|t|)>*u<around*|(|t|)>|]><rprime|'>=\<mu\><around*|(|t|)>*q<around*|(|t|)>\<Longrightarrow\>\<mu\><around*|(|t|)>*u<around*|(|t|)>=<big|int>\<mu\><around*|(|t|)>*q<around*|(|t|)>*\<mathd\>t+c
    </equation*>

    <\equation*>
      \<Longrightarrow\><with|color|blue|u<around*|(|t|)>=<frac|1|\<mu\><around*|(|t|)>>*<around*|[|<big|int>\<mu\><around*|(|t|)>*q<around*|(|t|)>*\<mathd\>t+c|]>>,
    </equation*>

    where

    <\equation*>
      <with|color|blue|\<mu\><around*|(|t|)>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t>>
    </equation*>

    is called the <strong|integrating factor>.

    <\example>
      Find the <strong|general solution> of

      <\equation*>
        <around*|(|4+t<rsup|2>|)>*<frac|d*y|d*t>+2*t*y=4*t
      </equation*>

      <strong|Answer:> Rewrite it in the standard form

      <\equation*>
        y<rprime|'>+<frac|2*t|4+t<rsup|2>>*y=<frac|4*t|4+t<rsup|2>>.
      </equation*>

      Here <math|p<around*|(|t|)>=<frac|2*t|4+t<rsup|2>>>. Then find the
      integrating factor <math|\<mu\><around*|(|t|)>>:

      <\equation*>
        \<mu\><around*|(|t|)>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t>=e<rsup|ln
        <around*|(|4+t<rsup|2>|)>>=4+t<rsup|2>.
      </equation*>

      So the solution is

      <\equation*>
        y<around*|(|t|)>=<frac|1|4+t<rsup|2>>*<around*|(|<big|int><around*|(|4+t<rsup|2>|)>*<frac|4*t|4+t<rsup|2>>*\<mathd\>t+c|)>=<frac|1|4+t<rsup|2>>*<around*|(|2*t<rsup|2>+c|)>
      </equation*>
    </example>

    <\example>
      Solve the <strong|initial value problem>

      <gather*|<tformat|<table|<row|<cell|t*y<rprime|'>+2*y=4*t<rsup|2>>>|<row|<cell|<text|initial
      condition: >y<around|(|1|)>=2>>>>>

      <strong|Answer:> First rewrite the equation into the standard form:

      <\equation*>
        y<rprime|'>+<frac|2|t>*y=4*t.
      </equation*>

      Find the integrating factor:

      <\equation*>
        \<mu\>=e<rsup|<big|int><frac|2|t>*\<mathd\>t>=e<rsup|2*ln
        <around*|\||t|\|>>=e<rsup|ln t<rsup|2>>=t<rsup|2>.
      </equation*>

      The general solution is

      <\equation*>
        y=<frac|1|t<rsup|2>>*<around*|[|<big|int>
        t<rsup|2>*4*t*\<mathd\>t+c|]>=<frac|1|t<rsup|2>>*<around*|[|t<rsup|4>+c|]>=t<rsup|2>+<frac|c|t<rsup|2>>.
      </equation*>

      Plugging the initial condition:

      <\equation*>
        y<around*|(|1|)>=2\<Longrightarrow\>1+<frac|1|c>=2\<Longrightarrow\><with|color|blue|c=1>.
      </equation*>

      The solution of the initial value problem is

      <\equation*>
        y=t<rsup|2>+<frac|1|t<rsup|2>>.
      </equation*>
    </example>

    \;

    <section|Separable equations>

    <\example>
      Solve

      <\equation*>
        <frac|d*y|d*x>=<frac|x<rsup|2>|1-y<rsup|2>>.
      </equation*>

      This equation is nonlinear. We can separate the variables <math|x> and
      <math|y> as follows

      <\equation*>
        <around*|(|1-y<rsup|2>|)>*\<mathd\>y=x<rsup|2>*\<mathd\>x\<Longrightarrow\><big|int><around*|(|1-y<rsup|2>|)>*\<mathd\>y=<big|int>x<rsup|2>*\<mathd\>x
      </equation*>

      <\equation*>
        \<Longrightarrow\>y-<frac|y<rsup|3>|3>=<frac|x<rsup|3>|3>+c
      </equation*>

      This is an example of <strong|implicit solutions>.
    </example>

    <\definition>
      An ODE in the form of

      <\equation*>
        <frac|\<mathd\>y|\<mathd\>x>=<frac|f<around*|(|x|)>|g<around*|(|y|)>>
      </equation*>

      is called <strong|separable.>
    </definition>

    We can solve it as follows

    <\equation*>
      <big|int>f<around*|(|x|)>*\<mathd\>x=<big|int>g<around*|(|y|)>*\<mathd\>y
    </equation*>

    <\example>
      Solve the initial value problem

      <\equation*>
        <frac|d*y|d*x>=<frac|3*x<rsup|2>+4*x+2|2*<around|(|y-1|)>>,<space|1em>y<around|(|0|)>=-1
      </equation*>

      <strong|Solution:>\ 

      <\equation*>
        <big|int><around*|(|3*x<rsup|2>+4*x+2|)>*\<mathd\>x=<big|int>2*<around*|(|y-1|)>*\<mathd\>y
      </equation*>

      <\equation*>
        \<Rightarrow\><space|1em>x<rsup|3>+2*x<rsup|2>+2*x=y<rsup|2>-2*y+c.
      </equation*>

      Plugging the initial condition

      <\equation*>
        y<around*|(|0|)>=-1\<Rightarrow\>0=1+2+c\<Rightarrow\>c=-3.
      </equation*>

      The solution to the initial value problem is

      <\equation*>
        x<rsup|3>+2*x<rsup|2>+2*x=y<rsup|2>-2*y-3.
      </equation*>

      <\question>
        What is the domain and range of the solution?
      </question>
    </example>

    <\example>
      Recall the differential equation for continuous compound interests:

      <\equation*>
        u<rprime|'>=r*u.
      </equation*>

      Note this equation is both linear and separable. As a separable
      equation, we have

      <\equation*>
        <frac|\<mathd\>u|\<mathd\>t>=r*u\<Rightarrow\><frac|\<mathd\>u|u>=r*\<mathd\>t\<Rightarrow\><big|int><frac|\<mathd\>u|u>=<big|int>r*\<mathd\>t\<Rightarrow\>ln
        <around*|\||u|\|>=r*t+c
      </equation*>

      <\equation*>
        \<Rightarrow\><around*|\||u|\|>=c*e<rsup|r*t>\<Rightarrow\>u=c*e<rsup|r*t>.
      </equation*>

      If the initial condition is <math|u<around*|(|0|)>=u<rsub|0>>,then we
      find <math|c=u<rsub|0>>.

      <\exercise>
        Solve the equation by the method of integrating factor.
      </exercise>
    </example>

    <section|Exact Equations>

    <subsection|Motivation and definition>

    Suppose <math|\<psi\><around*|(|x,y|)>=c> is an solution of some ODE.
    Taking <math|d/\<mathd\>x> on both sides of the solution.

    <\equation*>
      <frac|d|\<mathd\>x>*\<psi\><around*|(|x,y|)>=<frac|d|\<mathd\>x>*c\<Rightarrow\><with|color|blue|<frac|\<partial\>\<psi\>|\<partial\>x>>+<with|color|dark
      green|<frac|\<partial\>\<psi\>|\<partial\>y>>*<frac|\<mathd\>y|\<mathd\>x>=0\<Rightarrow\><with|color|blue|M<around*|(|x,y|)>>+<with|color|dark
      green|N<around*|(|x,y|)>>*y<rprime|'>=0,
    </equation*>

    where

    <\equation*>
      M<around*|(|x,y|)>=\<partial\><rsub|x>\<psi\>,<space|1em>N<around*|(|x,y|)>=\<partial\><rsub|y>*\<psi\>.
    </equation*>

    \;

    <\example>
      Solve <math|<with|color|blue|2*x+y<rsup|2>>+<with|color|dark
      green|2*x*y>*y<rprime|'>=0>.

      <\answer*>
        Guess the solution. Let <math|\<psi\>=x<rsup|2>+y<rsup|2>*x>. Then

        <\equation*>
          \<psi\><rsub|x>=2*x+y<rsup|2>,<space|1em>\<psi\><rsub|y>=2*x*y.
        </equation*>

        So

        <\equation*>
          0=\<psi\><rsub|x>+\<psi\><rsub|y>*y<rprime|'>=<frac|d|\<mathd\>x>*\<psi\><around*|(|x,y|)>
        </equation*>

        So the solution is

        <\equation*>
          \<psi\><around*|(|x,y|)>=c.
        </equation*>
      </answer*>
    </example>

    <\definition*>
      An ODE of the form

      <\equation*>
        M<around|(|x,y|)>+N<around|(|x,y|)>*y<rprime|'>=0<infix-or>M<around*|(|x,y|)>*\<mathd\>x+N<around*|(|x,y|)>*\<mathd\>y=0
      </equation*>

      is called <strong|exact> if there exists
      <math|\<psi\><around*|(|x,y|)>> such that

      <\equation*>
        \<psi\><rsub|x>=M,<space|1em>\<psi\><rsub|y>=N.
      </equation*>

      The solution of the equation is

      <\equation*>
        \<psi\><around*|(|x,y|)>=c,
      </equation*>

      where <math|c> is an arbitrary constant.
    </definition*>

    <subsection|Theorem and method>

    <\theorem>
      Suppose an ODE can be written in the form

      <\equation>
        M<around|(|x,y|)>+N<around|(|x,y|)>*y<rprime|'>=0<infix-or>M<around*|(|x,y|)>*\<mathd\>x+N<around*|(|x,y|)>*\<mathd\>y=0
      </equation>

      where the functions <math|M,N,M<rsub|y>> and <math|N<rsub|x>> are all
      continuous in the rectangular region
      <math|R=<around*|[|a,b|]>\<times\><around*|[|c,d|]>>. Then Eq. (1) is
      an exact differential equation <strong|if and only if>

      <\equation*>
        M<rsub|y><around|(|x,y|)>=N<rsub|x><around|(|x,y|)>,\<forall\><around|(|x,y|)>\<in\>R.
      </equation*>
    </theorem>

    <\proof>
      <math|<rprime|''>\<Longrightarrow\><rprime|''>>. Suppose Eq. (1) is
      exact. Then there exists a <math|\<psi\><around*|(|x,y|)>> such that

      <\equation*>
        \<psi\><rsub|x>=M,<space|1em>\<psi\><rsub|y>=N.
      </equation*>

      Then

      <\equation*>
        M<rsub|y>=\<psi\><rsub|x*y>,<space|1em>N<rsub|x>=\<psi\><rsub|y*x>.
      </equation*>

      Since <math|M<rsub|y>,N<rsub|x>> are continuous, we have
      <math|\<psi\><rsub|x*y>> and <math|\<psi\><rsub|y*x>> are continuous.
      So\ 

      <\equation*>
        \<psi\><rsub|x*y>=\<psi\><rsub|y*x>.
      </equation*>

      i.e.

      <\equation*>
        M<rsub|y>=N<rsub|x>.
      </equation*>

      <math|<rprime|''>\<Longleftarrow\><rprime|''>> Suppose
      <math|M<rsub|y>=N<rsub|x>>. We want to find a function
      <math|\<psi\><around*|(|x,y|)>> such that <math|\<psi\><rsub|x>=M> and
      <math|\<psi\><rsub|y>=N>. Let

      <\equation*>
        \<psi\>=<big|int>M<around*|(|x,y|)>*\<mathd\>x+h<around*|(|y|)>.
      </equation*>

      Then <math|\<psi\><rsub|x>=M>, and

      <\equation*>
        \<psi\><rsub|y>=\<partial\><rsub|y><big|int>M<around*|(|x,y|)>*\<mathd\>x+h<rprime|'><around*|(|y|)>.
      </equation*>

      We want <math|\<psi\><rsub|y>=N>, that is

      <\equation*>
        h<rprime|'><around*|(|y|)>=N<around*|(|x,y|)>-\<partial\><rsub|y><big|int>M<around*|(|x,y|)>*\<mathd\>x.
      </equation*>

      We need the RHS to be independent of <math|x>. That is

      <\equation*>
        <frac|\<partial\>|\<partial\>x> <around*|[|N<around*|(|x,y|)>-\<partial\><rsub|y>
        <big|int>*M<around*|(|x,y|)>*\<mathd\>x|]>=0.
      </equation*>

      Let's check:

      <\eqnarray*>
        <tformat|<table|<row|<cell|<frac|\<partial\>|\<partial\>x><around*|[|N<around*|(|x,y|)>-\<partial\><rsub|y><big|int>*M<around*|(|x,y|)>*\<mathd\>x|]>>|<cell|=>|<cell|N<rsub|x>-\<partial\><rsub|y>
        \<partial\><rsub|x><big|int>M*\<mathd\>x=N<rsub|x>-M<rsub|y>=0.>>>>
      </eqnarray*>

      \;
    </proof>

    \;

    <\example>
      Solve the ODE

      <\equation*>
        <around*|(|y*cos x+2*x*e<rsup|y>|)>+<around*|(|sin
        x+x<rsup|2>*e<rsup|y>-1|)>*y<rprime|'>=0.
      </equation*>

      <strong|Answer:>\ 

      <\eqnarray*>
        <tformat|<table|<row|<cell|M<rsub|y>>|<cell|=>|<cell|cos
        x+2*x*e<rsup|y>>>|<row|<cell|N<rsub|x>>|<cell|=>|<cell|cos
        x+2*x*e<rsup|y>>>>>
      </eqnarray*>

      So <math|M<rsub|y>=N<rsub|x>>, and the equation is exact.

      Next, let

      <\equation*>
        \<psi\>=<big|int>M d*x=<big|int>y*<around*|(|cos x|)>+2*x*e<rsup|y>
        d*x=y*<around*|(|sin x|)>+x<rsup|2>*e<rsup|y>+h<around*|(|y|)>.
      </equation*>

      Then

      <\equation*>
        \<psi\><rsub|y>=sin x+x<rsup|2>*e<rsup|y>+h<rprime|'><around*|(|y|)>=N=sin
        x+x<rsup|2>*e<rsup|y>-1
      </equation*>

      <\equation*>
        \<Longrightarrow\><space|1em>h<rprime|'><around*|(|y|)>=-1<space|1em>\<Rightarrow\><space|1em>h<around*|(|y|)>=-y.
      </equation*>

      So the solution is

      <\equation*>
        \<psi\>=y*<around*|(|sin x|)>+x<rsup|2>*e<rsup|y>-y=c.
      </equation*>
    </example>

    <\exercise*>
      Solve the above equation, but using <math|\<psi\>=<big|int>N
      d*y+h<around*|(|x|)>> first.
    </exercise*>

    <\question*>
      What is the relationship between separable and exact equations?
    </question*>

    <subsection|Integrating factors>

    Sometimes we can multiply a function to a non-exact equation to make it
    exact. Take a function <math|\<mu\><around*|(|x,y|)>\<neq\>0>,\ 

    <\eqnarray*>
      <tformat|<table|<row|<cell|M<around*|(|x,y|)> d*x+N<around*|(|x,y|)>
      d*y>|<cell|=>|<cell|0>>|<row|<cell|\<mu\><around*|(|x,y|)><around*|[|M<around*|(|x,y|)>
      d*x+N<around*|(|x,y|)> d*y|]>>|<cell|=>|<cell|0>>|<row|<cell|\<mu\><around*|(|x,y|)>*M<around*|(|x,y|)>
      d*x+\<mu\><around*|(|x,y|)>*N<around*|(|x,y|)>
      d*y>|<cell|=>|<cell|0>>|<row|<cell|<wide|M|~><around*|(|x,y|)>
      d*x+<wide|N|~><around*|(|x,y|)> d*y>|<cell|=>|<cell|0>>>>
    </eqnarray*>

    where <math|<wide|M|~><around*|(|x,y|)>=\<mu\><around*|(|x,y|)>*M<around*|(|x,y|)>,<wide|N|~><around*|(|x,y|)>=\<mu\><around*|(|x,y|)>*N<around*|(|x,y|)>>.
    Then let

    <\equation*>
      <wide|M|~><rsub|y>=\<mu\><rsub|y>*M+\<mu\>*M<rsub|y>,<space|1em><wide|N|~><rsub|x>=\<mu\><rsub|x>*N+\<mu\>*N<rsub|x>.
    </equation*>

    We want <math|<wide|M|~><rsub|y>=<wide|N|~><rsub|x>>, i.e.

    <\equation*>
      \<mu\><rsub|y>*M+\<mu\>*M<rsub|y>=\<mu\><rsub|x>*N+\<mu\>*N<rsub|x>.
    </equation*>

    Let's choose <math|\<mu\>> such that <math|\<mu\><rsub|y>=0>. Then the
    above equation reduces to

    <\equation*>
      \<mu\>*M<rsub|y>=\<mu\><rsub|x>*N+\<mu\>*N<rsub|x><space|1em>\<Leftrightarrow\>\<mu\><rsub|x>=<frac|M<rsub|y>-N<rsub|x>|N>*\<mu\>.
    </equation*>

    If the function <math|<around*|(|M<rsub|y>-N<rsub|x>|)>/N> is a function
    of <math|x> only, then we can solve <math|\<mu\>> as a separable
    equation. Here <math|\<mu\>> is called an integrating factor.

    <\wide-block>
      <tformat|<table|<row|<\cell>
        <\example>
          Solve the ODE

          <\equation*>
            <around*|(|3*x*y+y<rsup|2>|)>+<around*|(|x<rsup|2>+x*y|)>*y<rprime|'>=0.
          </equation*>

          <strong|Answer:> It's first order, nonlinear, and not separable.
          Check if it's exact:

          <\equation*>
            M<rsub|y>=3*x+2*y,<space|1em>N<rsub|x>=2*x+y.
          </equation*>

          Not exact!. Next, try integrating factors.

          <\equation*>
            <frac|M<rsub|y>-N<rsub|x>|N>=<frac|x+y|x<rsup|2>+x*y>=<frac|1|x>
          </equation*>

          is a function of <math|x> only! Let

          <\equation*>
            \<mu\><rprime|'><around*|(|x|)>=<frac|1|x>*\<mu\><space|1em>\<Rightarrow\><space|1em>\<mu\><around*|(|x|)>=x.
          </equation*>

          Then multiply <math|\<mu\>> to the original equation:

          <\equation*>
            x*<around*|(|3*x*y+y<rsup|2>|)>+x*<around*|(|x<rsup|2>+x*y|)>*y<rprime|'>=0
          </equation*>

          <\equation*>
            <around*|(|3*x<rsup|2>*y+x*y<rsup|2>|)>+<around*|(|x<rsup|3>+x*<rsup|2>y|)>*y<rprime|'>=0
          </equation*>

          Double check the new equation is exact! Then solve it as usual
          (exercise).
        </example>
      </cell>>>>
    </wide-block>

    Similarly, if <math|<around*|(|N<rsub|x>-M<rsub|y>|)>/M> is a function of
    <math|y> only, then we can use the integrating factor
    <math|\<mu\><around*|(|y|)>> solving

    <\equation*>
      \<mu\><rprime|'>=<frac|N<rsub|x>-M<rsub|y>|M>*\<mu\>.
    </equation*>

    <section|Direction fields>

    Consider the first order ODE:

    <\equation*>
      y<rprime|'>=f<around*|(|t,y|)>
    </equation*>

    Draw small arrows as a vector <math|<around*|(|1,f<around*|(|t,y|)>|)>>
    at many points <math|<around*|(|t,y|)>>

    Online plotter:

    <slink|https://aeb019.hosted.uark.edu/dfield.html>

    <\example*>
      Consider

      <\equation*>
        y<rprime|'>=<frac|y*cos x|1+3*y<rsup|3>>
      </equation*>
    </example*>

    \;

    <section|The Existence and Uniqueness Theorem>

    <subsection|Linear equations>

    <\theorem*>
      Consider the initial value problem

      <\equation*>
        y<rprime|'>+p<around*|(|t|)>*y=q<around*|(|t|)>,<space|1em>y<around*|(|t<rsub|0>|)>=y<rsub|0>.
      </equation*>

      If <math|p,q> are continuous on an interval <math|I=<around*|[|a,b|]>>
      containing <math|t<rsub|0>>, then the IVP has a unique solution on
      <math|I>.
    </theorem*>

    <\example*>
      Consider

      <\equation*>
        t*y<rprime|'>+2*y=4*t<rsup|2>,<space|1em>y<around|(|1|)>=2.
      </equation*>

      Solve it by integrating factors,\ 

      <\equation*>
        y<rprime|'>+<frac|2|t>*y=4*t<space|1em>\<Rightarrow\><space|1em>\<mu\>=exp<around*|[|<big|int><frac|2|t>
        d*t|]>=t<rsup|2>.
      </equation*>

      <\equation*>
        y=<frac|1|t<rsup|2>><around*|[|<big|int>4*t<rsup|3>
        d*t+c|]>=<frac|1|t<rsup|2>><around*|[|t<rsup|4>+c|]>=t<rsup|2>+<frac|c|t<rsup|2>>.
      </equation*>

      Plugging <math|y<around*|(|1|)>=2>, we obtain <math|c=1>. The solution
      is

      <\equation*>
        y=t<rsup|2>+<frac|1|t<rsup|2>>.
      </equation*>

      Now, <math|p<around*|(|t|)>=<frac|2|t>,q<around*|(|t|)>=4*t>. So
      <math|p,q> are continuous in <math|<around*|(|-\<infty\>,0|)>\<cup\><around*|(|0,\<infty\>|)>>.
      But <math|1\<in\><around*|(|0,\<infty\>|)>> only, so we know from the
      theorem the IVP has a unique solution in
      <math|<around*|(|0,\<infty\>|)>>, which is

      <\equation*>
        y=t<rsup|2>+<frac|1|t<rsup|2>>,<space|1em>t\<in\><around*|(|0,\<infty\>|)>.
      </equation*>

      If the initial condition is changed to <math|y<around*|(|-1|)>=2>, then
      the solution is

      <\equation*>
        y=t<rsup|2>+<frac|1|t<rsup|2>>,<space|1em>t\<in\><around*|(|-\<infty\>,0|)>.
      </equation*>
    </example*>

    <subsection|Nonlinear equations>

    <\theorem*>
      Consider the initial value problem

      <\equation*>
        y<rprime|'>=f<around*|(|t,y|)>,<space|1em>y<around*|(|t<rsub|0>|)>=y<rsub|0>.
      </equation*>

      If <math|f> and <math|\<partial\><rsub|y>f> are continuous on a
      rectangular domain <math|R=<around*|[|a,b|]>\<times\><around*|[|c,d|]>>
      containing the point <math|<around*|(|t<rsub|0>,y<rsub|0>|)>>. Then the
      IVP has a unique solution in some interval <math|I> containing
      <math|t<rsub|0>>.
    </theorem*>

    <\example*>
      Consider the IVP.

      <\equation*>
        <frac|d*y|d*x>=<frac|3*x<rsup|2>+4*x+2|2*<around|(|y-1|)>>,<space|1em>y<around|(|0|)>=-1
      </equation*>

      It is separable. Let's solve it first,\ 

      <\equation*>
        2<around*|(|y-1|)>*d*y=<around*|(|3*x<rsup|2>+4*x+2|)>*d*x<space|1em>\<Rightarrow\><space|1em>y<rsup|2>-2*y=x<rsup|3>+2*x<rsup|2>+2*x+c
      </equation*>

      <\equation*>
        y<around*|(|0|)>=-1<space|1em>\<Rightarrow\><space|1em>c=3.
      </equation*>

      The solution is

      <\equation*>
        y<rsup|2>-2*y=x<rsup|3>+2*x<rsup|2>+2*x+3
      </equation*>

      <\equation*>
        y=<frac|2\<pm\><sqrt|4+4<around*|(|x<rsup|3>+2*x<rsup|2>+2*x+3|)>>|2>=1\<pm\><sqrt|x<rsup|3>+2*x<rsup|2>+2*x+4>=1-<sqrt|x<rsup|3>+2*x<rsup|2>+2*x+4>.
      </equation*>

      Here <math|f> and <math|\<partial\><rsub|y> f> are continuous
      everywhere except <math|y=1>.
    </example*>

    <\example*>
      Consider

      <\equation*>
        y<rprime|'>=y<rsup|1/3>,<space|1em>y<around|(|0|)>=0*<space|1em><around|(|t\<geq\>0|)>
      </equation*>

      First, let's solve it as a separable equation.

      <\equation*>
        y<rsup|-1/3> d*y=d*t<space|1em>\<Rightarrow\><space|1em><frac|3|2>*y<rsup|2/3>=t+c
      </equation*>

      Plugging <math|y<around*|(|0|)>=0> yields <math|c=0>. So

      <\equation*>
        y=\<pm\><around*|(|<frac|2|3>*t|)><rsup|3/2>
      </equation*>

      are two solutions. In addition

      <\equation*>
        y=0
      </equation*>

      is also a solution. In fact, we have infinitely many solutions defined
      as

      <\equation*>
        y=<choice|<tformat|<table|<row|<cell|0,>|<cell|t\<less\>t<rsub|0>>>|<row|<cell|<around*|[|<around*|\<nobracket\>|<frac|2|3>*<around*|(||\<nobracket\>>*t-t<rsub|0>|)>|]><rsup|3/2>,>|<cell|t\<geqslant\>t<rsub|0>>>>>>,<infix-or>y=<choice|<tformat|<table|<row|<cell|0,>|<cell|t\<less\>t<rsub|0>>>|<row|<cell|-<around*|[|<frac|2|3>*<around*|(|t-t<rsub|0>|)>|]><rsup|3/2>,>|<cell|t\<geqslant\>t<rsub|0>>>>>>
      </equation*>

      for any <math|t<rsub|0>\<gtr\>0>. (<strong|Exercise:> check <math|y> is
      continuous and differentiable at <math|t=t<rsub|0>>.)

      <with|gr-mode|<tuple|edit|cline>|gr-frame|<tuple|scale|1cm|<tuple|0.5gw|0.5gh>>|gr-geometry|<tuple|geometry|1par|0.6par>|gr-color|red|gr-auto-crop|true|<graphics||<with|arrow-end|\<gtr\>|<line|<point|-4.57028|0.387382>|<point|4.96593484625935|0.342558096553603>>>|<with|arrow-end|\<gtr\>|<line|<point|-4.17808|-1.41677>|<point|-4.14446076476367|2.65097322062586>>>|<with|color|blue|<spline|<point|-4.16318|0.385468>|<point|-3.11422059653126|0.380537956827157>>>|<with|color|blue|<spline|<point|-3.13536|0.380637>|<point|-1.71278075542543|0.779588144314855>|<point|0.0241335369590382|2.02344289563534>>>|<with|color|blue|<spline|<point|-3.08258|0.380389>|<point|-1.24788053659304|0.371765435654606>>>|<with|color|blue|<spline|<point|-1.23317|0.371696>|<point|-0.0879254496464111|0.72355865101213>|<point|1.08868418236822|1.62003054385572>>>|<with|color|red|<spline|<point|-4.16318|0.385468>|<point|-3.50561163248595|0.382377470061263>>>|<with|color|red|<spline|<point|-3.48365|0.382274>|<point|-2.71011546355651|-0.0944719512076496>|<point|-2.23946771981362|-1.11420872931724>>>|<with|color|red|<spline|<point|-3.38995|0.381834>|<point|-1.2478805398271|0.371765435669807>>>|<with|color|red|<spline|<point|-1.24788|0.371765>|<point|-0.489782110416146|0.368202079416764>>>|<with|color|red|<spline|<point|-0.485879|0.368184>|<point|0.449957686059745|0.129646022003249>|<point|1.33521395290021|-0.408237113702907>>>|<with|color|red|<point|-2.7672|1.19743>>>>
    </example*>

    In fact,\ 

    <\equation*>
      f=y<rsup|1/3>,<space|1em>\<partial\><rsub|y> f=<frac|1|3>*y<rsup|-2/3>.
    </equation*>

    So <math|\<partial\><rsub|y> f> is discontinuous near
    <math|<around*|(|0,0|)>>. So there exists no rectangle <math|R>
    containing <math|<around*|(|0,0|)>> such that <math|f,\<partial\><rsub|y>
    f> are both continuous in <math|R>. So we can't gaurantee the existence
    and uniqueness of solution for the IVP.

    <\note>
      One may not be able to find all solutions to nonlinear equations using
      one method.\ 
    </note>

    <\example*>
      Consider

      <\equation*>
        <frac|d*y|d*t>=y<rsup|2>,<space|1em>y<around*|(|0|)>=y<rsub|0>\<neq\>0
      </equation*>

      We can first solve it as a separable eqn.

      <\equation*>
        y<rsup|-2>*d*y=d*t<space|1em>\<Rightarrow\><space|1em>-y<rsup|-1>=t+c<space|1em>\<Rightarrow\><space|1em>y=-<frac|1|t+c><space|1em>\<Rightarrow\><space|1em>y=-<frac|1|t-<frac|1|y<rsub|0>>>.
      </equation*>

      Now <math|f=y<rsup|2>,\<partial\><rsub|y> f=2*y> are continuous
      everywhere. However, the solution is not defined for every <math|t>.
      For example, if <math|y<rsub|0>\<gtr\>0>,then the solution is defined
      only in <math|<around*|(|-\<infty\>,<frac|1|y<rsub|0>>|)>>.
    </example*>

    <section|Applications>

    <subsection|Falling object in the air>

    <\equation*>
      m*v<rprime|'>=m*g-\<gamma\>*v,
    </equation*>

    where <math|v> is the velocity, <math|m,g,\<gamma\>> are constants.\ 

    <\itemize>
      <item>Analyze the solutions using direction field.

      <item>Solve it by integrating factors.

      <\equation*>
        v<rprime|'>+<frac|\<gamma\>|m>*v=g
      </equation*>

      integrating factor

      <\equation*>
        \<mu\>=e<rsup|<big|int>\<gamma\>/m>=e<rsup|<frac|\<gamma\>|m>*t>
      </equation*>

      <\equation*>
        v<around*|(|t|)>=e<rsup|-<frac|\<gamma\>|m>*t>*<around*|[|<big|int>g*e<rsup|<frac|\<gamma\>|m>
        t>*d*t+c|]>=e<rsup|-<frac|\<gamma\>|m>*t>*<around*|[|<frac|g*m|\<gamma\>>*e<rsup|<frac|\<gamma\>|m>*t>+c|]>=<frac|g*m|\<gamma\>>+c*e<rsup|-<frac|\<gamma\>|m>*t>.
      </equation*>

      If the initial condition is <math|v<around*|(|0|)>=v<rsub|0>>. Then
      <math|c=v<rsub|0>-g*m/\<gamma\>>. So the solution of the IVP is

      <\equation*>
        v<around*|(|t|)>=<frac|g*m|\<gamma\>>+<around*|[|v<rsub|0>-<frac|g*m|\<gamma\>>|]>*e<rsup|-<frac|\<gamma\>|m>*t>.
      </equation*>

      So

      <\equation*>
        lim<rsub|t\<rightarrow\>\<infty\>>
        v<around*|(|t|)>=<frac|g*m|\<gamma\>>.
      </equation*>

      All other solutions converge to the <strong|equilibrium solution>
      <math|v=g*m/\<gamma\>> as <math|t\<rightarrow\>\<infty\>>. This
      equilibrium solution is a <strong|stable> one.
    </itemize>

    <subsection|Compound interest with deposits/withdrawals>

    Assume the annual interest rate is <math|r>. The continuous rate of
    deposit/withdrawal is <math|k>. Then the ODE model for the total balance
    <math|u<around*|(|t|)>> is

    <\equation*>
      u<rprime|'>=r*u+k.
    </equation*>

    integrating factor

    <\equation*>
      \<mu\>=e<rsup|-r*t>
    </equation*>

    <\equation*>
      u=e<rsup|r*t>*<around*|[|<big|int>k*e<rsup|-r*t>+c|]>=e<rsup|r*t>*<around*|[|-<frac|k|r>*e<rsup|-r*t>+c|]>=-<frac|k|r>+c*e<rsup|r*t>.
    </equation*>

    If the initial condition is <math|u<around*|(|0|)>=u<rsub|0>>, then
    <math|c=u<rsub|0>+k/r>. So the solution of the IVP is

    <\equation*>
      u=-<frac|k|r>+<around*|(|u<rsub|0>+<frac|k|r>|)>*e<rsup|r*t>.
    </equation*>

    The equilibrium solution is <math|u=-<frac|k|r>>, and it is an
    <strong|unstable> one since all other solutions diverge from it as
    <math|t\<rightarrow\>\<infty\>>.

    <subsection|Population dynamics>

    <subsubsection|Exponential growth>

    <\equation*>
      y<rprime|'>=r*y,
    </equation*>

    The solution is

    <\equation*>
      y=y<rsub|0>*e<rsup|r*t>
    </equation*>

    where <math|y<rsub|0>=y<around*|(|0|)>>.\ 

    <\itemize>
      <item>If <math|r\<gtr\>0>, we have exponential growth

      <item>If <math|r\<less\>0>, we have exponential decay, such as
      radioactive decay.
    </itemize>

    <subsubsection|Logistic growth>

    <\equation*>
      y<rprime|'>=<around*|(|r-a*y|)>*y.
    </equation*>

    Note that the right-hand-side depends on <math|y> only. In general, ODE
    of the form

    <\equation*>
      y<rprime|'>=f<around*|(|y|)>
    </equation*>

    is called <strong|autonomous>. There are two equilibrium solutions

    <\equation*>
      y=0,<space|1em>y=<frac|r|a>.
    </equation*>

    From the direction field we can tell the equilibrium solution <math|y=0>
    is unstable, while the solution <math|y=<frac|r|a>> is stable.

    <\wide-centered>
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    </wide-centered>

    Now let's solve the equation:

    <\equation*>
      <frac|d*y|<around*|(|r-a*y|)>*y>=d*t<space|1em>\<Rightarrow\><space|1em><big|int><frac|d*y|<around*|(|r-a*y|)>*y>=<big|int>d*t<space|1em>\<Rightarrow\>
    </equation*>

    <\equation*>
      <frac|1|<around*|(|r-a*y|)>*y>=<frac|A|y>+<frac|B|r-a*y>=<frac|A*<around*|(|r-a*y|)>+B*y|y*<around*|(|r-a*y|)>><space|1em>\<Rightarrow\><space|1em>1=A*<around*|(|r-a*y|)>+B*y
    </equation*>

    <\equation*>
      y=0\<Rightarrow\>A=1/r,<space|1em>y=r/a\<Rightarrow\>B=a/r
    </equation*>

    So

    <\equation*>
      <big|int><frac|d*y|<around*|(|r-a*y|)>*y>=<big|int><frac|1|r*y>+<frac|a/r|r-a*y>*d*y=<frac|1|r>*ln
      <around*|\||y|\|>+<frac|a|r><around*|(|<frac|1|-a>|)>*ln
      <around*|\||r-a*y|\|>=<frac|1|r>*ln <around*|\||y|\|>-<frac|1|r>*ln
      <around*|\||r-a*y|\|>
    </equation*>

    <\equation*>
      =<frac|1|r>*ln <frac|<around*|\||y|\|>|<around*|\||r-a*y|\|>>=<big|int>d*t=t+c
    </equation*>

    <\equation*>
      \<Rightarrow\><space|1em><frac|<around*|\||y|\|>|<around*|\||r-a*y|\|>>=e<rsup|r*<around*|(|t+c|)>>=c*e<rsup|r*t><space|1em>\<Rightarrow\><frac|y|r-a*y>=c*e<rsup|r*t>.
    </equation*>

    <\equation*>
      \<Rightarrow\><space|1em>y=<frac|r*c*e<rsup|r*t>|1+a*c*e<rsup|r*t>>=<frac|r*c|e<rsup|-r*t>+a*c>=<frac|r|<frac|1|c>*e<rsup|-r*t>+a>.
    </equation*>

    Suppose the initial condition is <math|y<around*|(|0|)>=y<rsub|0>>, then\ 

    <\equation*>
      c=<frac|y<rsub|0>|r-a*y<rsub|0>><space|1em>\<Rightarrow\><space|1em>y=<frac|r|<frac|r-a*y<rsub|0>|y<rsub|0>>*e<rsup|-r*t>+a>=<frac|r*y<rsub|0>|a*y<rsub|0>+<around*|(|r-a*y<rsub|0>|)>*e<rsup|-r*t>>=<block|<tformat|<table|<row|<cell|<frac|K*y<rsub|0>|y<rsub|0>+<around*|(|K-y<rsub|0>|)>*e<rsup|-r*t>>>>>>>,
    </equation*>

    where <math|K=<frac|r|a>>. Note that <math|y<rprime|'>=<around*|(|r-a*y|)>*y=r*<around*|(|1-<frac|y|K>|)>*y>

    <\enumerate>
      <item>If <math|0\<less\>y<rsub|0>\<less\>K>, then
      <math|y<around*|(|t|)>> is an increasing function, and
      <math|lim<rsub|t\<rightarrow\>\<infty\>> y<around*|(|t|)>=K>, but
      <math|y<around*|(|t|)>\<less\>K> for all <math|t\<gtr\>0>. Moreover,
      <math|lim<rsub|t\<rightarrow\>\<infty\>> y<rprime|'><around*|(|t|)>=0>,
      and

      <\equation*>
        y<rprime|''>=<frac|d|d*t> <frac|d*y|d*t>=<frac|d|d*t>
        f<around*|(|y|)>=f<rprime|'><around*|(|y|)>*<frac|d*y|d*t>=f<rprime|'><around*|(|y|)>*f<around*|(|y|)>,
      </equation*>

      where

      <\equation*>
        f<around*|(|y|)>=r*<around*|(|1-<frac|y|K>|)>*y,<space|1em>f<rprime|'><around*|(|y|)>=r*<around*|(|1-<frac|2*y|K>|)>
      </equation*>

      <\enumerate>
        <item>If <math|0\<less\>y\<less\><frac|K|2>>, then
        <math|y<rprime|''>\<gtr\>0>, so the graph is concave up.

        <item>If <math|<frac|K|2>\<less\>y\<less\>K>, then
        <math|y<rprime|''>\<less\>0>, so the graph is concave down.
      </enumerate>

      <item>If <math|y<rsub|0>\<gtr\>K>, then <math|y<around*|(|t|)>> is an
      decreasing function, and <math|lim<rsub|t\<rightarrow\>\<infty\>>
      y<around*|(|t|)>=K>, but <math|y<around*|(|t|)>\<gtr\>K> for all
      <math|t\<gtr\>0>. Moreover, <math|lim<rsub|t\<rightarrow\>\<infty\>>
      y<rprime|'><around*|(|t|)>=0>, and <math|y<rprime|''><around*|(|t|)>\<gtr\>0>
      for all <math|t>.
    </enumerate>

    <section|Euler's method>

    Consider a general 1st order ODE

    <\equation*>
      y<rprime|'>=f<around*|(|t,y|)>.
    </equation*>

    Take <math|<around*|(|t<rsub|0>,y<rsub|0>|)>>, then

    <\equation*>
      y<rprime|'><around*|(|t<rsub|0>|)>=f<around*|(|t<rsub|0>,y<rsub|0>|)>
    </equation*>

    <\equation*>
      y<rprime|'><around*|(|t<rsub|0>|)>=lim<rsub|h\<rightarrow\>0>
      <frac|y<around*|(|t<rsub|0>+h|)>-y<around*|(|t<rsub|0>|)>|h>\<approx\><frac|y<around*|(|t<rsub|1>|)>-y<around*|(|t<rsub|0>|)>|t<rsub|1>-t<rsub|0>>
    </equation*>

    if <math|<around*|\||t<rsub|1>-t<rsub|0>|\|>> is small. So

    <\equation*>
      y<around*|(|t<rsub|1>|)>\<approx\>y<around*|(|t<rsub|0>|)>+<around*|(|t<rsub|1>-t<rsub|0>|)>*y<rprime|'><around*|(|t<rsub|0>|)>=y<around*|(|t<rsub|0>|)>+<around*|(|t<rsub|1>-t<rsub|0>|)>*f<around*|(|t<rsub|0>,y<rsub|0>|)>
    </equation*>

    Let

    <\equation*>
      y<rsub|1>=y<rsub|0>+<around*|(|t<rsub|1>-t<rsub|0>|)>*f<around*|(|t<rsub|0>,y<rsub|0>|)>
    </equation*>

    So <math|y<rsub|1>\<approx\>y<around*|(|t<rsub|1>|)>>. Repeat this
    process, we obtain an algorithm: For a sequence of
    <math|t<rsub|0>,t<rsub|1>,t<rsub|2>,\<ldots\>>

    <\equation*>
      <block|<tformat|<table|<row|<cell|y<rsub|k+1>=y<rsub|k>+<around*|(|t<rsub|k+1>-t<rsub|k>|)>*f<around*|(|t<rsub|k>,y<rsub|k>|)>>>>>>
    </equation*>

    This sequence of <math|y<rsub|0>,y<rsub|1>,y<rsub|2>,\<ldots\>> is an
    approximation of the true values <math|y<around*|(|t<rsub|0>|)>,y<around*|(|t<rsub|1>|)>,y<around*|(|t<rsub|2>|)>,\<ldots\>>

    \;

    \;

    \;
  </shown>>
</body>

<\initial>
  <\collection>
    <associate|eqn-ver-sep|<macro|0.0fn>>
    <associate|font-base-size|10>
    <associate|info-flag|minimal>
    <associate|magnification|1.7>
    <associate|page-medium|papyrus>
    <associate|par-first|1tab>
    <associate|par-hyphen|professional>
    <associate|par-kerning-margin|true>
    <associate|preamble|false>
  </collection>
</initial>

<\references>
  <\collection>
    <associate|auto-1|<tuple|?|?>>
    <associate|auto-10|<tuple|5.1|?>>
    <associate|auto-11|<tuple|5.2|?>>
    <associate|auto-12|<tuple|6|?>>
    <associate|auto-13|<tuple|6.1|?>>
    <associate|auto-14|<tuple|6.2|?>>
    <associate|auto-15|<tuple|6.3|?>>
    <associate|auto-16|<tuple|6.3.1|?>>
    <associate|auto-17|<tuple|6.3.2|?>>
    <associate|auto-18|<tuple|7|?>>
    <associate|auto-2|<tuple|1|?>>
    <associate|auto-3|<tuple|2|?>>
    <associate|auto-4|<tuple|3|?>>
    <associate|auto-5|<tuple|3.1|?>>
    <associate|auto-6|<tuple|3.2|?>>
    <associate|auto-7|<tuple|3.3|?>>
    <associate|auto-8|<tuple|4|?>>
    <associate|auto-9|<tuple|5|?>>
  </collection>
</references>

<\auxiliary>
  <\collection>
    <\associate|toc>
      <vspace*|2fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|font-size|<quote|1.19>|Chapter
      2: First Order Differential Equations>
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-1><vspace|1fn>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|1<space|2spc>Method
      of integrating factors> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-2><vspace|0.5fn>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|2<space|2spc>Separable
      equations> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-3><vspace|0.5fn>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|3<space|2spc>Exact
      Equations> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-4><vspace|0.5fn>

      <with|par-left|<quote|1tab>|3.1<space|2spc>Motivation and definition
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-5>>

      <with|par-left|<quote|1tab>|3.2<space|2spc>Theorem and method
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-6>>

      <with|par-left|<quote|1tab>|3.3<space|2spc>Integrating factors
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-7>>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|4<space|2spc>Direction
      fields> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-8><vspace|0.5fn>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|5<space|2spc>The
      Existence and Uniqueness Theorem> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-9><vspace|0.5fn>

      <with|par-left|<quote|1tab>|5.1<space|2spc>Linear equations
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-10>>

      <with|par-left|<quote|1tab>|5.2<space|2spc>Nonlinear equations
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-11>>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|6<space|2spc>Applications>
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-12><vspace|0.5fn>

      <with|par-left|<quote|1tab>|6.1<space|2spc>Falling object in the air
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-13>>

      <with|par-left|<quote|1tab>|6.2<space|2spc>Compound interest with
      deposits/withdrawals <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-14>>

      <with|par-left|<quote|1tab>|6.3<space|2spc>Population dynamics
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-15>>

      <with|par-left|<quote|2tab>|6.3.1<space|2spc>Exponential growth
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-16>>

      <with|par-left|<quote|2tab>|6.3.2<space|2spc>Logistic growth
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-17>>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|7<space|2spc>Euler's
      method> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-18><vspace|0.5fn>
    </associate>
  </collection>
</auxiliary>