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A differential equation is an equation between specified derivatives of a function, its 0 [6 i( X0 w; P9 L
valves,and known quantities.Many laws of physics are most simply and naturally formu- + M4 R. t2 `4 i( P g0 ^7 m9 S
lated as differential equations (or DE’s, as we shall write for short).For this reason,DE’s
3 J1 Q5 h! R8 u( \, I5 P" {have been studies by the greatest mathematicians and mathematical physicists since the 1 p; g7 W1 H% s' O# t
time of Newton.. " o( b/ N2 r# e5 l
Ordinary differential equations are DE’s whose unknowns are functions of a single va-
* g" T9 W- c+ D1 vriable;they arise most commonly in the study of dynamic systems and electric networks.
# W- O- r1 |* V8 x7 n2 p( cThey are much easier to treat than partial differential equations,whose unknown functions
0 `* {8 ~# A8 a8 e& z+ x1 I) idepend on two or more independent variables. 9 ?- {$ o% I& C5 E: e0 c3 I" E
Ordinary DE’s are classified according to their order. The order of a DE is defined as 2 |8 f4 Y) ?8 k+ H
the largest positive integer, n, for which an n-th derivative occurs in the equation. This
! m4 E0 n/ x7 q0 t% j! achapter will be restricted to real first order DE’s of the form
* B9 H8 C& l, q# T4 H Φ(x, y, y′)=0 (1) " O( f- \: M; C( Y
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. + i2 T Y0 q. j) R0 _
DEFINITION A solution of (1)is a differentiable function f(x) such that
, w% ^8 T- _3 A4 P3 }9 u* yΦ(x. f(x),f′(x))=0 for all x in the interval where f(x) is defined. ) U5 X7 U7 Q6 f2 [) q( K I6 _6 x
EXAMPLE 1. In the first-other DE 7 ?5 W/ J# s5 d, d( B+ A; }
x+yy′=0 (2)
2 N6 e2 r8 o3 a$ R# V$ B- z, Qthe function Φ is a polynomial function Φ(x, y, z)=x+ yz of three variables in- 9 B c9 C9 [3 u
volved. The solutions of (2) can be found by considering the identity
+ J/ d! l, d8 n; z* n9 t9 o$ Dd(x²+y²)/d x=2(x+yyˊ).From this identity,one sees that x²+y² is a con-
4 _- o9 r5 c4 u e/ w# `stant if y=f(x) is any solution of (2). ) d! }# m0 T A$ s
The equation x²+y²=c defines y implicitly as a two-valued function of x,
) ]% c; c! F+ n1 g: V: X: D/ bfor any positive constant c.Solving for y,we get two solutions,the(single-valued)
# m5 k. Y" O% Nfunctions y=±(c-x²)0.5 ,for each positive constant c.The graphs of these so- $ \) \9 W5 V" t6 f
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.
+ L/ c; b( ^% `On the x-axis,where y=0,the DE(2) implies that x=0.Hence the DE has no solutions 2 E4 l& Y$ b: Q7 p
which cross the x-axis,except possibly at the origin.This fact is easily overlooked, ' u4 s4 \/ D0 Q6 g5 s( h" a% V' P3 T
because the solution curves appear to cross the x-axis;hence yˊdoes not exist,and the DE (2) is not satisfied there. / x; g; H$ _1 t; Z7 Y
The 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. - ]6 L: H$ `3 u* |; I$ f3 H$ _
DEFINITION. A normal first-order DE is one of the form $ j$ ~. e, b" X4 W7 a" }& F" O
yˊ=F(x,y) (3)
3 w' T9 a9 {5 C: l' i' ]4 KIn 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. 7 a: l- D2 n3 W8 y m
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: J g( u; A; d2 P) [3 z Fundamental Theorem of the Calculus. , q( ]7 E+ f, y6 c# B8 b+ n) |- q9 O
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/ `$ v$ ]' F& c: |" ^ The most familiar class of differential equations consists of the first-order DE’s of the form + e. g+ s+ r5 J- W: v, o. {
yˊ=g(x) (4) ) y9 ?2 F5 G* G7 B4 `
Such DE’s are normal and their solutions are descried by the fundamental thorem of the calculus,which reads as follows. ) y% ^* r- r5 ~; a+ m. T0 g
FUNDAMENTAL 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 / \1 B( I6 Y. C4 b/ M4 c, k3 H
f(x)=c+∫axg(t)dt , c=f(a) (5) & \) w- o0 v; ?' g( ~; h
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. 5 h+ r" ~3 q- \, k" ]( Z
We recall that the definite integral
0 R, B& X& z( M S/ p∫axg(t)dt=lim(maxΔtk->0)Σg(tk)Δtk , Δtk=tk-tk-1 (5ˊ) % o2 |, L$ M0 h
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. & |! @, F5 H6 d' h9 @$ ?# _: k
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Solutions and Integrals " j& k9 ` U5 X. H- E, a4 w
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; w# m+ d R6 p# a2 ^6 } 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.
; R5 m8 L( k" iDEFINITION. 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. & T( T: R# r: j @5 V
In 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. ' K; D& p) [2 n2 d/ F1 U( x; D }
The second-order DE " _+ w! G9 R8 y4 F$ Q
d²x/dt²=-x (2ˊ)
1 \! m8 P. L. `* vbecomes a first-order DE equivalent to (2) after setting dx/dx=y:
( X j0 [. a5 Ty ( dy/dx )=-x (2)
/ r6 V& R$ {4 vAs we have seen, the curves u(x,y)=x²+y²=c are integrals of this DE.When the DE (2ˊ) 1 ^8 h4 C3 z0 W& y. l
is interpreted as equation of motion under Newton’s second law,the integrals
, o/ n1 \+ s% `' k2 ^% Xc=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. 6 ~) e1 p5 O( k6 W
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