mirror of https://gitee.com/XmacsLabs/planet.git
1177 lines
101 KiB
Plaintext
1177 lines
101 KiB
Plaintext
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<TeXmacs|2.1.2>
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<style|<tuple|beamer|invisible-multiply|reddish|framed-theorems>>
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<\body>
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<screens|<\shown>
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<doc-data|<doc-title|Ordinary Differential
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Equations>|<doc-author|<author-data|<author-name|Yuliang Wang>>>>
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<chapter*|Chapter 2: First Order Differential Equations>
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General form\
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<\equation*>
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F<around*|(|u,u<rprime|'>|)>=0
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</equation*>
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<\example>
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<dueto|falling object in air, general solution, integral curves,
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initial value problem>
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The motion of the object is governed by the Newton's law. Let
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<math|v<around*|(|t|)>> be the velocity of the object at time <math|t>.
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<\equation*>
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m*v<rprime|'>=m*g-\<gamma\>*v,
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</equation*>
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where <math|m> is the mass of the object, <math|g> is the gravitational
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constant, and <math|\<gamma\>> is the coefficient of air resistant
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force.
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This is a first order linear ODE.
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</example>
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<section|Method of integrating factors>
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<\example>
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Find the general solutions of
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<\equation*>
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u<rprime|'>+u=2
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</equation*>
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idea: combine <math|u<rprime|'>+u> into the derivative of another
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function.\
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Consider the product rule of differentiation:
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<\equation*>
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<around*|(|f*g|)><rprime|'>=f<rprime|'>*g+f*g<rprime|'>.
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</equation*>
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Let <math|f=u<around*|(|t|)>,g=e<rsup|t>>, then
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<\equation*>
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<around*|[|u<around*|(|t|)>*e<rsup|t>|]><rprime|'>=u<rprime|'>*e<rsup|t>+u*e<rsup|t>=e<rsup|t>*<around*|(|u<rprime|'>+u|)>.
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</equation*>
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Now, we multiply the original equation by <math|e<rsup|t>>:
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<\equation*>
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<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
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</equation*>
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</example>
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Divide by <math|e<rsup|t>>:
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<\equation*>
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u<around*|(|t|)>=e<rsup|-t>*<around*|[|2*e<rsup|t>+c|]>=2+c*e<rsup|-t>.
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</equation*>
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<strong|Method of integrating factors:>
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Consider the 1st order linear ODE (standard form):
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<\equation*>
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<with|color|dark green|u<rprime|'><around*|(|t|)>+p<around*|(|t|)>*u<around*|(|t|)>=q<around*|(|t|)>.>
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</equation*>
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Multiply left side by <math|\<mu\><around*|(|t|)>>:
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<\equation*>
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<around*|[|u<rprime|'><around*|(|t|)>+p<around*|(|t|)>*u<around*|(|t|)>|]>*\<mu\><around*|(|t|)>=u<rprime|'>*\<mu\>+p*u*\<mu\>.
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</equation*>
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We want
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<\equation*>
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\<mu\>=g,p*\<mu\>=g<rprime|'>
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</equation*>
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i.e.
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<\equation*>
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\<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>>.
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</equation*>
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Check (by the chain rule and fundamental theorem of calculus)
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<\equation*>
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\<mu\><rprime|'>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t>*p<around*|(|t|)>=\<mu\><around*|(|t|)>*p<around*|(|t|)>.
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</equation*>
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Derivation directly:
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<\equation*>
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\<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|)>
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</equation*>
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<\equation*>
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\<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.>
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</equation*>
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So the original ODE becomes
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<\equation*>
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<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
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</equation*>
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<\equation*>
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\<Longrightarrow\><with|color|blue|u<around*|(|t|)>=<frac|1|\<mu\><around*|(|t|)>>*<around*|[|<big|int>\<mu\><around*|(|t|)>*q<around*|(|t|)>*\<mathd\>t+c|]>>,
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</equation*>
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where
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<\equation*>
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<with|color|blue|\<mu\><around*|(|t|)>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t>>
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</equation*>
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is called the <strong|integrating factor>.
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<\example>
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Find the <strong|general solution> of
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<\equation*>
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<around*|(|4+t<rsup|2>|)>*<frac|d*y|d*t>+2*t*y=4*t
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</equation*>
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<strong|Answer:> Rewrite it in the standard form
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<\equation*>
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y<rprime|'>+<frac|2*t|4+t<rsup|2>>*y=<frac|4*t|4+t<rsup|2>>.
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</equation*>
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Here <math|p<around*|(|t|)>=<frac|2*t|4+t<rsup|2>>>. Then find the
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integrating factor <math|\<mu\><around*|(|t|)>>:
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<\equation*>
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\<mu\><around*|(|t|)>=e<rsup|<big|int>p<around*|(|t|)>*\<mathd\>t>=e<rsup|ln
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<around*|(|4+t<rsup|2>|)>>=4+t<rsup|2>.
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</equation*>
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So the solution is
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<\equation*>
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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|)>
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</equation*>
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</example>
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<\example>
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Solve the <strong|initial value problem>
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<gather*|<tformat|<table|<row|<cell|t*y<rprime|'>+2*y=4*t<rsup|2>>>|<row|<cell|<text|initial
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condition: >y<around|(|1|)>=2>>>>>
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<strong|Answer:> First rewrite the equation into the standard form:
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<\equation*>
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y<rprime|'>+<frac|2|t>*y=4*t.
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</equation*>
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Find the integrating factor:
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<\equation*>
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\<mu\>=e<rsup|<big|int><frac|2|t>*\<mathd\>t>=e<rsup|2*ln
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<around*|\||t|\|>>=e<rsup|ln t<rsup|2>>=t<rsup|2>.
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</equation*>
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The general solution is
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<\equation*>
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y=<frac|1|t<rsup|2>>*<around*|[|<big|int>
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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>>.
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</equation*>
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Plugging the initial condition:
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<\equation*>
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y<around*|(|1|)>=2\<Longrightarrow\>1+<frac|1|c>=2\<Longrightarrow\><with|color|blue|c=1>.
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</equation*>
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The solution of the initial value problem is
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<\equation*>
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y=t<rsup|2>+<frac|1|t<rsup|2>>.
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</equation*>
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</example>
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\;
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<section|Separable equations>
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<\example>
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Solve
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<\equation*>
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<frac|d*y|d*x>=<frac|x<rsup|2>|1-y<rsup|2>>.
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</equation*>
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This equation is nonlinear. We can separate the variables <math|x> and
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<math|y> as follows
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<\equation*>
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<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
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</equation*>
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<\equation*>
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\<Longrightarrow\>y-<frac|y<rsup|3>|3>=<frac|x<rsup|3>|3>+c
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</equation*>
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This is an example of <strong|implicit solutions>.
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</example>
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<\definition>
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An ODE in the form of
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<\equation*>
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<frac|\<mathd\>y|\<mathd\>x>=<frac|f<around*|(|x|)>|g<around*|(|y|)>>
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</equation*>
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is called <strong|separable.>
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</definition>
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We can solve it as follows
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<\equation*>
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<big|int>f<around*|(|x|)>*\<mathd\>x=<big|int>g<around*|(|y|)>*\<mathd\>y
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</equation*>
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<\example>
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Solve the initial value problem
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<\equation*>
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<frac|d*y|d*x>=<frac|3*x<rsup|2>+4*x+2|2*<around|(|y-1|)>>,<space|1em>y<around|(|0|)>=-1
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</equation*>
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<strong|Solution:>\
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<\equation*>
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<big|int><around*|(|3*x<rsup|2>+4*x+2|)>*\<mathd\>x=<big|int>2*<around*|(|y-1|)>*\<mathd\>y
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</equation*>
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<\equation*>
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\<Rightarrow\><space|1em>x<rsup|3>+2*x<rsup|2>+2*x=y<rsup|2>-2*y+c.
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</equation*>
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Plugging the initial condition
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<\equation*>
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y<around*|(|0|)>=-1\<Rightarrow\>0=1+2+c\<Rightarrow\>c=-3.
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</equation*>
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The solution to the initial value problem is
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<\equation*>
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x<rsup|3>+2*x<rsup|2>+2*x=y<rsup|2>-2*y-3.
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</equation*>
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<\question>
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What is the domain and range of the solution?
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</question>
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</example>
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<\example>
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Recall the differential equation for continuous compound interests:
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<\equation*>
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u<rprime|'>=r*u.
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</equation*>
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Note this equation is both linear and separable. As a separable
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equation, we have
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<\equation*>
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<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
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<around*|\||u|\|>=r*t+c
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</equation*>
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<\equation*>
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\<Rightarrow\><around*|\||u|\|>=c*e<rsup|r*t>\<Rightarrow\>u=c*e<rsup|r*t>.
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</equation*>
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If the initial condition is <math|u<around*|(|0|)>=u<rsub|0>>,then we
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find <math|c=u<rsub|0>>.
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<\exercise>
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Solve the equation by the method of integrating factor.
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</exercise>
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</example>
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<section|Exact Equations>
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<subsection|Motivation and definition>
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Suppose <math|\<psi\><around*|(|x,y|)>=c> is an solution of some ODE.
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Taking <math|d/\<mathd\>x> on both sides of the solution.
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<\equation*>
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<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
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green|<frac|\<partial\>\<psi\>|\<partial\>y>>*<frac|\<mathd\>y|\<mathd\>x>=0\<Rightarrow\><with|color|blue|M<around*|(|x,y|)>>+<with|color|dark
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green|N<around*|(|x,y|)>>*y<rprime|'>=0,
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</equation*>
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where
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<\equation*>
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M<around*|(|x,y|)>=\<partial\><rsub|x>\<psi\>,<space|1em>N<around*|(|x,y|)>=\<partial\><rsub|y>*\<psi\>.
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</equation*>
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\;
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<\example>
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Solve <math|<with|color|blue|2*x+y<rsup|2>>+<with|color|dark
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green|2*x*y>*y<rprime|'>=0>.
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<\answer*>
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Guess the solution. Let <math|\<psi\>=x<rsup|2>+y<rsup|2>*x>. Then
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<\equation*>
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\<psi\><rsub|x>=2*x+y<rsup|2>,<space|1em>\<psi\><rsub|y>=2*x*y.
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</equation*>
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So
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<\equation*>
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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>
|
||
|
|
<image|<tuple|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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>>
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<no-break><pageref|auto-5>>
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<with|par-left|<quote|1tab>|3.2<space|2spc>Theorem and method
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
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<no-break><pageref|auto-6>>
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<with|par-left|<quote|1tab>|3.3<space|2spc>Integrating factors
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
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<no-break><pageref|auto-7>>
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<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|4<space|2spc>Direction
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fields> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
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<no-break><pageref|auto-8><vspace|0.5fn>
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<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|5<space|2spc>The
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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>>
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<no-break><pageref|auto-9><vspace|0.5fn>
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<with|par-left|<quote|1tab>|5.1<space|2spc>Linear equations
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
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<no-break><pageref|auto-10>>
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<with|par-left|<quote|1tab>|5.2<space|2spc>Nonlinear equations
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
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<no-break><pageref|auto-11>>
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<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|6<space|2spc>Applications>
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
|
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<no-break><pageref|auto-12><vspace|0.5fn>
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<with|par-left|<quote|1tab>|6.1<space|2spc>Falling object in the air
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
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<no-break><pageref|auto-13>>
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<with|par-left|<quote|1tab>|6.2<space|2spc>Compound interest with
|
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deposits/withdrawals <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
|
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<no-break><pageref|auto-14>>
|
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<with|par-left|<quote|1tab>|6.3<space|2spc>Population dynamics
|
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
|
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<no-break><pageref|auto-15>>
|
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<with|par-left|<quote|2tab>|6.3.1<space|2spc>Exponential growth
|
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|
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
|
||
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<no-break><pageref|auto-16>>
|
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<with|par-left|<quote|2tab>|6.3.2<space|2spc>Logistic growth
|
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<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
|
||
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<no-break><pageref|auto-17>>
|
||
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<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|7<space|2spc>Euler's
|
||
|
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method> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
|
||
|
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<no-break><pageref|auto-18><vspace|0.5fn>
|
||
|
|
</associate>
|
||
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</collection>
|
||
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</auxiliary>
|