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< ><FONT face="Times New Roman" size=3>A differential equation is an equation between specified derivatives of a function, its</FONT></P>
< ><FONT face="Times New Roman" size=3>valves,and known quantities.Many laws of physics are most simply and naturally formu-</FONT></P>
< ><FONT face="Times New Roman" size=3>lated as differential equations (or DE’s, as we shall write for short).For this reason,DE’s </FONT></P>
<P ><FONT face="Times New Roman" size=3>have been studies by the greatest mathematicians and mathematical physicists since the </FONT></P>
<P ><FONT face="Times New Roman" size=3>time of Newton..</FONT></P>
<P ><FONT face="Times New Roman" size=3>Ordinary differential equations are DE’s whose unknowns are functions of a single va-</FONT></P>
<P ><FONT face="Times New Roman" size=3>riable;they arise most commonly in the study of dynamic systems and electric networks.</FONT></P>
<P ><FONT face="Times New Roman" size=3>They are much easier to treat than partial differential equations,whose unknown functions</FONT></P>
<P ><FONT face="Times New Roman" size=3>depend on two or more independent variables.</FONT></P>
<P ><FONT face="Times New Roman" size=3>Ordinary DE’s are classified according to their order. The order of a DE is defined as </FONT></P>
<P ><FONT face="Times New Roman" size=3>the largest positive integer, n, for which an n-th derivative occurs in the equation. This</FONT></P>
<P ><FONT face="Times New Roman" size=3>chapter will be restricted to real first order DE’s of the form</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman"> </FONT>Φ<FONT face="Times New Roman">(x, y, y</FONT>′<FONT face="Times New Roman">)=0 (1)</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">Given the function </FONT>Φ<FONT face="Times New Roman">of three real variables, the problem is to determine all real functions y=f(x) which satisfy the DE, that is ,all solutions of(1)in the following sense.</FONT></FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3><B>DEFINITION </B> A solution of (1)is a differentiable function f(x) such that </FONT></FONT></P>
<P ><FONT size=3>Φ<FONT face="Times New Roman">(x. f(x),f</FONT>′<FONT face="Times New Roman">(x))=0 for all x in the interval where f(x) is defined.</FONT></FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3><B>EXAMPLE 1.</B> In the first-other DE</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman"> x+yy</FONT>′<FONT face="Times New Roman">=0 (2)</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">the function </FONT>Φ<FONT face="Times New Roman"> is a polynomial function </FONT>Φ<FONT face="Times New Roman">(x, y, z)=x+ yz of three variables in-</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>volved. The solutions of (2) can be found by considering the identity</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">d(x</FONT>²<FONT face="Times New Roman">+y</FONT>²<FONT face="Times New Roman">)/d x=2(x+yy</FONT>ˊ<FONT face="Times New Roman">).From this identity,one sees that x</FONT>²<FONT face="Times New Roman">+y</FONT>²<FONT face="Times New Roman"> is a con-</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>stant if y=f(x) is any solution of (2).</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">The equation x</FONT>²<FONT face="Times New Roman">+y</FONT>²<FONT face="Times New Roman">=c defines y implicitly as a two-valued function of x,</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>for any positive constant c.Solving for y,we get two solutions,the(single-valued)</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">functions y=</FONT>±<FONT face="Times New Roman">(c-x</FONT>²<FONT face="Times New Roman">)<SUP>0.5</SUP> ,for each positive constant c.The graphs of these so-</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>lutions,the so-called solution curves,form two families of scmicircles,which fill the upper half-plane y>0 and the lower half-plane y>0,respectively.</FONT></P>
<P ><FONT face="Times New Roman" size=3>On the x-axis,where y=0,the DE(2) implies that x=0.Hence the DE has no solutions</FONT></P>
<P ><FONT face="Times New Roman" size=3>which cross the x-axis,except possibly at the origin.This fact is easily overlooked,</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">because the solution curves appear to cross the x-axis;hence y</FONT>ˊ<FONT face="Times New Roman">does not exist,and the DE (2) is not satisfied there.</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">The preceding difficulty also arises if one tries to solve the DE(2)for y</FONT>ˊ<FONT face="Times New Roman">. Dividing through by y,one gets y</FONT>ˊ<FONT face="Times New Roman">=-x/y,an equation which cannot be satisfied if y=0.The preceding difficulty is thus avoided if one restricts attention to regions where the DE(1) is normal,in the following sense.</FONT></FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3><B> DEFINITION. </B> A normal first-order DE is one of the form</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman"> y</FONT>ˊ<FONT face="Times New Roman">=F(x,y) (3)</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">In the normal form y</FONT>ˊ<FONT face="Times New Roman">=-x/y of the DE (2),the function F(x,y) is continuous in the upper half-plane y>0 and in the lower half-plane where y<0;it is undefined on the x-axis.</FONT></FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3> <p></p></FONT></FONT></P>
<P ><B><FONT face="Times New Roman"><FONT size=3>Fundamental Theorem of the Calculus.<p></p></FONT></FONT></B></P>
<P ><FONT face="Times New Roman"><FONT size=3> <p></p></FONT></FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3>The most familiar class of differential equations consists of the first-order DE’s of the form </FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman"> y</FONT>ˊ<FONT face="Times New Roman">=g(x) (4)</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>Such DE’s are normal and their solutions are descried by the fundamental thorem of the calculus,which reads as follows.</FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3><B>FUNDAMENTAL THEOREM OF THE CALCULUS</B>. Let the function g(x)in DE(4) be continuous in the interval a<x<b.Given a number c,there is one and only one solution f(x) of the DE(4) in the interval such that f(a)=c. This solution is given by the definite integral</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">f(x)=c+</FONT>∫<FONT face="Times New Roman"><SUB>a</SUB><SUP>x</SUP>g(t)dt , c=f(a) (5)</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>This basic result serves as a model of rigorous formulation in several respects. First,it specifies the region under consideration,as a vertical strip a<x<b in the xy-plane.Second,it describes in precise terms the class of functions g(x) considered.And third, it asserts the existence and uniqueness of a solution,given the “initial condition”f(a)=c.</FONT></P>
<P ><FONT face="Times New Roman" size=3>We recall that the definite integral </FONT></P>
<P ><FONT size=3>∫<FONT face="Times New Roman"><SUB>a</SUB><SUP>x</SUP>g(t)dt=lim(max</FONT>Δ<FONT face="Times New Roman">t<SUB>k</SUB>->0)</FONT>Σ<FONT face="Times New Roman">g(t<SUB>k</SUB>)</FONT>Δ<FONT face="Times New Roman">t<SUB>k </SUB>, </FONT>Δ<FONT face="Times New Roman">t<SUB>k</SUB>=t<SUB>k</SUB>-t<SUB>k-1 </SUB> (5</FONT>ˊ<FONT face="Times New Roman">)</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">is defined for each fixed x as a limit of Ricmann sums; it is not necessary to find a formal expression for the indefinite integral </FONT>∫<FONT face="Times New Roman"> g(x) dx to give meaning to the definite integral </FONT>∫<FONT face="Times New Roman"><SUB>a</SUB><SUP>x</SUP>g(t)dt,provided only that g(t) is continuous.Such functions as the error function crf x =(2/(</FONT>π<FONT face="Times New Roman">)<SUP>0.5</SUP>)</FONT>∫<FONT face="Times New Roman"><SUB>0</SUB><SUP>x</SUP>e<SUP>-t</SUP></FONT><SUP>²<FONT face="Times New Roman"> </FONT></SUP><FONT face="Times New Roman">dt and the sine integral function SI(x)=</FONT>∫<SUB><FONT face="Times New Roman">x</FONT></SUB><SUP>∞</SUP><FONT face="Times New Roman">[(sin t )/t]dt are indeed commonly defined as definite integrals.</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman"> <p></p></FONT></FONT></P>
<P align=center><B><FONT face="Times New Roman">Solutions and Integrals<p></p></FONT></B></P>
<P ><FONT face="Times New Roman"><FONT size=3> <p></p></FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">According to the definition given above a solution of a DE is always a function. For example, the solutions of the DE x+yy</FONT>ˊ<FONT face="Times New Roman">=0 in Example I are the functions y=</FONT>±<FONT face="Times New Roman"> (c-x</FONT>²<FONT face="Times New Roman">)<SUP>0.5</SUP>,whose graphs are semicircles of arbitrary diameter,centered at the origin.The graph of the solution curves are ,however,more easily described by the equation x</FONT>²<FONT face="Times New Roman">+y</FONT>²<FONT face="Times New Roman">=c,describing a family of circles centered at the origin.In what sense can such a family of curves be considered as a solution of the DE ?To answer this question,we require a new notion.</FONT></FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3><B>DEFINITION.</B> An integral of DE(1)is a function of two variables,u(x,y),which assumes a constant value whenever the variable y is replaced by a solution y=f(x) of the DE.</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">In the above example, the function u(x,y)=x</FONT>²<FONT face="Times New Roman">+y</FONT>²<FONT face="Times New Roman"> is an integral of the DE x+yy</FONT>ˊ<FONT face="Times New Roman"> =0,because,upon replacing the variable y by any function </FONT>±<FONT face="Times New Roman">( c-x</FONT>²<FONT face="Times New Roman">)<SUP>0.5</SUP>,we obtain u(x,y)=c.</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>The second-order DE</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman"> d</FONT>²<FONT face="Times New Roman">x/dt</FONT>²<FONT face="Times New Roman">=-x (2</FONT>ˊ<FONT face="Times New Roman">)</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>becomes a first-order DE equivalent to (2) after setting dx/dx=y:</FONT></P>
<P ><FONT face="Times New Roman" size=3>y ( dy/dx )=-x (2)</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">As we have seen, the curves u(x,y)=x</FONT>²<FONT face="Times New Roman">+y</FONT>²<FONT face="Times New Roman">=c are integrals of this DE.When the DE (2</FONT>ˊ<FONT face="Times New Roman">)</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>is interpreted as equation of motion under Newton’s second law,the integrals </FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">c=x</FONT>²<FONT face="Times New Roman">+y</FONT>²<FONT face="Times New Roman"> represent curves of constant energy c.This illustrates an important principle:an integral of a DE representing some kind of motion is a quantity that remains unchanged through the motion.</FONT></FONT></P>
<P ><FONT size=3><FONT face="Times New Roman"> <p></p></FONT></FONT></P> |
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