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A differential equation is an equation between specified derivatives of a function, its 8 d, h( |4 Z# v( X% Q4 t B5 v$ n! ?
valves,and known quantities.Many laws of physics are most simply and naturally formu- 9 U7 M. s& i2 s2 H5 k$ B
lated as differential equations (or DE’s, as we shall write for short).For this reason,DE’s , F3 r4 G# s9 r
have been studies by the greatest mathematicians and mathematical physicists since the
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Ordinary differential equations are DE’s whose unknowns are functions of a single va-
9 r( c; h8 l! O3 A4 T, a& E8 eriable;they arise most commonly in the study of dynamic systems and electric networks.
' E! G, b! x$ K3 c6 mThey are much easier to treat than partial differential equations,whose unknown functions 1 _$ n/ |/ ?4 W+ g) N2 b& O
depend on two or more independent variables. & E6 n. X8 [/ x5 W$ ]( V4 r
Ordinary DE’s are classified according to their order. The order of a DE is defined as
' a1 `- O' i0 E+ W0 T# }the largest positive integer, n, for which an n-th derivative occurs in the equation. This 2 y$ y; x, z1 t# o5 t1 ]
chapter will be restricted to real first order DE’s of the form
( g8 x! l* c5 J* f i Φ(x, y, y′)=0 (1) ( [: l% B* }; D; j z
Given the function Φof 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.
* R: ?' l* C' _' ]+ [DEFINITION A solution of (1)is a differentiable function f(x) such that 3 s, K3 o O4 Y; t
Φ(x. f(x),f′(x))=0 for all x in the interval where f(x) is defined. ; |( Z+ _+ q" O" Z+ ~/ l* [* t; K& c( W
EXAMPLE 1. In the first-other DE % d; k% D9 l- C% k, i) T6 }
x+yy′=0 (2)
& i8 w# q0 S# T$ z% Lthe function Φ is a polynomial function Φ(x, y, z)=x+ yz of three variables in- " v0 | K. T3 h) b) N8 |! ?
volved. The solutions of (2) can be found by considering the identity . U$ {0 I' n$ _
d(x²+y²)/d x=2(x+yyˊ).From this identity,one sees that x²+y² is a con-
: ]+ p: G" Q& O$ ~% w# A0 I/ tstant if y=f(x) is any solution of (2). 8 X: w" V: C( N: P( c
The equation x²+y²=c defines y implicitly as a two-valued function of x,
D8 W; V* l. V' |for any positive constant c.Solving for y,we get two solutions,the(single-valued) 2 g6 V# w3 F3 s2 c6 o0 M. T
functions y=±(c-x²)0.5 ,for each positive constant c.The graphs of these so- ( h a3 o0 U7 k# E6 _4 S
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.
$ s! S4 \* }, H0 jOn the x-axis,where y=0,the DE(2) implies that x=0.Hence the DE has no solutions
. B9 [7 |" M5 g: L2 O, r$ Kwhich cross the x-axis,except possibly at the origin.This fact is easily overlooked, 7 d% [) a; l) y, P+ Y/ y. P
because the solution curves appear to cross the x-axis;hence yˊdoes not exist,and the DE (2) is not satisfied there.
/ s. \/ c7 ?4 ]* S6 m- g7 P; z3 @0 nThe preceding difficulty also arises if one tries to solve the DE(2)for yˊ. Dividing through by y,one gets yˊ=-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.
/ r4 n J7 y5 `8 ] DEFINITION. A normal first-order DE is one of the form
$ Y9 R: w0 U) G$ K3 X- E4 [9 y! K yˊ=F(x,y) (3)
( h8 W9 R/ e1 z' gIn the normal form yˊ=-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. 6 U8 e. x9 Z6 V3 \4 f' |* A z
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3 ~; J5 j' {, z% F* ^ Fundamental Theorem of the Calculus. + i( @5 x' x, j, C/ E/ k" x
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" n) ?# G4 L$ R8 x: i The most familiar class of differential equations consists of the first-order DE’s of the form % }* y! }/ w4 w0 A2 G
yˊ=g(x) (4)
( t0 g0 J" a, y) P$ t4 j8 X/ jSuch DE’s are normal and their solutions are descried by the fundamental thorem of the calculus,which reads as follows.
8 V) s/ ?, P' I! ~% yFUNDAMENTAL THEOREM OF THE CALCULUS. 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
, @5 m( ] N: t) gf(x)=c+∫axg(t)dt , c=f(a) (5)
" |4 r6 n/ |) k' dThis 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.
, f( k' Q1 _: o, C% IWe recall that the definite integral * N7 u! k8 ?% H1 t8 A
∫axg(t)dt=lim(maxΔtk->0)Σg(tk)Δtk , Δtk=tk-tk-1 (5ˊ) - l1 D4 w* A+ g
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 ∫ g(x) dx to give meaning to the definite integral ∫axg(t)dt,provided only that g(t) is continuous.Such functions as the error function crf x =(2/(π)0.5)∫0xe-t² dt and the sine integral function SI(x)=∫x∞[(sin t )/t]dt are indeed commonly defined as definite integrals.
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Solutions and Integrals
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6 d) s1 p: ]' ^' U; y# ~7 W. N/ v According to the definition given above a solution of a DE is always a function. For example, the solutions of the DE x+yyˊ=0 in Example I are the functions y=± (c-x²)0.5,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 x²+y²=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.
, N9 E( O' x) e6 Q* O4 [% s5 G7 A1 lDEFINITION. 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.
# O; M: w! E! l3 d- j, QIn the above example, the function u(x,y)=x²+y² is an integral of the DE x+yyˊ =0,because,upon replacing the variable y by any function ±( c-x²)0.5,we obtain u(x,y)=c. 2 ]% Z# D# o: ]6 l8 x) n+ W
The second-order DE
( ~( _" p$ I) D. B; F4 v1 s d²x/dt²=-x (2ˊ) 6 L/ x' I* T% z3 M
becomes a first-order DE equivalent to (2) after setting dx/dx=y:
! I( m9 R3 G# T' ]3 a2 m, \/ ey ( dy/dx )=-x (2)
% ^- a3 [( q0 T# @1 r8 hAs we have seen, the curves u(x,y)=x²+y²=c are integrals of this DE.When the DE (2ˊ) & X |& j! d1 r7 ^9 _4 g
is interpreted as equation of motion under Newton’s second law,the integrals $ S% v& u t* c4 ~/ X+ [* ]) _8 R
c=x²+y² 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.
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