THEORY AND PROBLEMS OF DIFFERENTIAL EQUATIONS
CHAPTER BASIC CONCEPTS DIFFERENTIAL EQUATIONS NOTATION SOLUTIONS INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS AN INTRODUCTION TO MODELING AND QUALITATIVE METHODS MATHEMATICAL MODELS THE “MODELING CYCLE” QUALITATIVE METHODS CLASSIFICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS STANDARD FORM AND DIFFERENTIAL FORM LINEAR EQUATIONS BERNOULLI EQUATIONS HOMOGENEOUS EQUATIONS SEPARABLE EQUATIONS EXACT EQUATIONS SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS GENERAL SOLUTION SOLUTIONS TO THE INITIAL-VALUE PROBLEM REDUCTION OF HOMOGENEOUS EQUATIONS EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS DEFINING PROPERTIES METHOD OF SOLUTION INTEGRATING FACTORS LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS REDUCTION OF BERNOULLI EQUATIONS CONTENTS CHAPTER APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS GROWTH AND DECAY PROBLEMS TEMPERATURE PROBLEMS FALLING BODY PROBLEMS DILUTION PROBLEMS ELECTRICAL CIRCUITS ORTHOGONAL TRAJECTORIES LINEAR DIFFERENTIAL EQUATIONS: THEORY OF SOLUTIONS LINEAR DIFFERENTIAL EQUATIONS LINEARLY INDEPENDENT SOLUTIONS THE WRONSKIAN NONHOMOGENEOUS EQUATIONS SECOND-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS INTRODUCTORY REMARK THE CHARACTERISTIC EQUATION THE GENERAL SOLUTION NTH-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS THE METHOD OF UNDETERMINED COEFFICIENTS SIMPLE FORM OF THE METHOD GENERALIZATIONS MODIFICATIONS LIMITATIONS OF THE METHOD VARIATION OF PARAMETERS THE METHOD SCOPE OF THE METHOD INITIAL-VALUE PROBLEMS FOR LINEAR DIFFERENTIAL EQUATIONS APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS SPRING PROBLEMS ELECTRICAL CIRCUIT PROBLEMS BUOYANCY PROBLEMS CLASSIFYING SOLUTIONS MATRICES MATRICES AND VECTORS MATRIX ADDITION SCALAR AND MATRIX MULTIPLICATION POWERS OF A SQUARE MATRIX DIFFERENTIATION AND INTEGRATION OF MATRICES EAT DEFINITION COMPUTATION OF EAT REDUCTION OF LINEAR DIFFERENTIAL EQUATIONS TO A SYSTEM OF FIRST-ORDER EQUATIONS AN EXAMPLE REDUCTION OF AN NTH ORDER EQUATION REDUCTION OF A SYSTEM GRAPHICAL AND NUMERICAL METHODS FOR SOLVING FIRST-ORDER DIFFERENTIAL EQUATIONS DIRECTION FIELDS EULER’S METHOD STABILITY FURTHER NUMERICAL METHODS FOR SOLVING FIRST-ORDER DIFFERENTIAL EQUATIONS GENERAL REMARKS MODIFIED EULER’S METHOD RUNGE–KUTTA METHOD ADAMS–BASHFORD–MOULTON METHOD MILNE’S METHOD STARTING VALUES ORDER OF A NUMERICAL METHOD NUMERICAL METHODS FOR SOLVING SECOND-ORDER DIFFERENTIAL EQUATIONS VIA SYSTEMS SECOND-ORDER DIFFERENTIAL EQUATIONS THE LAPLACE TRANSFORM PROPERTIES OF LAPLACE TRANSFORMS FUNCTIONS OF OTHER INDEPENDENT VARIABLES INVERSE LAPLACE TRANSFORMS MANIPULATING DENOMINATORS MANIPULATING NUMERATORS CONVOLUTIONS AND THE UNIT STEP FUNCTION CONVOLUTIONS UNIT STEP FUNCTION TRANSLATIONS SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS BY LAPLACE TRANSFORMS LAPLACE TRANSFORMS OF DERIVATIVES SOLUTIONS OF DIFFERENTIAL EQUATIONS SOLUTIONS OF LINEAR SYSTEMS BY LAPLACE TRANSFORMS SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS BY MATRIX METHODS SOLUTION OF THE INITIAL-VALUE PROBLEM SOLUTION WITH NO INITIAL CONDITIONS POWER SERIES SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS WITH VARIABLE COEFFICIENTS SECOND-ORDER EQUATIONS ANALYTIC FUNCTIONS AND ORDINARY POINTS SOLUTIONS AROUND THE ORIGIN OF HOMOGENEOUS EQUATIONS SOLUTIONS AROUND THE ORIGIN OF NONHOMOGENEOUS EQUATIONS INITIAL-VALUE PROBLEMS SOLUTIONS AROUND OTHER POINTS SERIES SOLUTIONS NEAR A REGULAR SINGULAR POINT REGULAR SINGULAR POINTS METHOD OF FROBENIUS SOME CLASSICAL DIFFERENTIAL EQUATIONS CLASSICAL DIFFERENTIAL EQUATIONS POLYNOMIAL SOLUTIONS AND ASSOCIATED CONCEPTS GAMMA AND BESSEL FUNCTIONS GAMMA FUNCTION BESSEL FUNCTIONS ALGEBRAIC OPERATIONS ON INFINITE SERIES AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS INTRODUCTORY CONCEPTS SOLUTIONS AND SOLUTION TECHNIQUES SECOND-ORDER BOUNDARY-VALUE PROBLEMS STANDARD FORM EIGENVALUE PROBLEMS STURM–LIOUVILLE PROBLEMS PROPERTIES OF STURM–LIOUVILLE PROBLEMS EIGENFUNCTION EXPANSIONS PIECEWISE SMOOTH FUNCTIONS FOURIER SINE SERIES FOURIER COSINE SERIES AN INTRODUCTION TO DIFFERENCE EQUATIONS INTRODUCTION CLASSIFICATIONS SOLVING DIFFERENTIAL EQUATIONS USING MATHEMATICA SOLVING SYSTEMS OF DIFFERENTIAL EQUATIONS VIA EIGENVALUES USING MATHEMATICA TERMINOLOGY EULER’S METHOD USING MICROSOFT EXCEL® SOME INTERESTING MODELING PROBLEMS EXAMPLES BASIC CONCEPTS DIFFERENTIAL EQUATIONS A DIFFERENTIAL EQUATION IS AN EQUATION INVOLVING AN UNKNOWN FUNCTION AND ITS DERIVATIVES EXAMPLE 1 1 THE FOLLOWING ARE DIFFERENTIAL EQUATIONS INVOLVING THE UNKNOWN FUNCTION Y DY DX X = + (1 1) E D Y DX DY DX Y + ? ?? ? ??= (1 2) D Y DX X D Y DX XY + + = (SIN ) (1 3) D Y DX Y DY DX Y DY DX X ? ? ? ? ? ?+ ? ?? ? ??+ ? ?? ? ??= (1 4) ? ? ? ? ? = Y T Y X (1 5) A DIFFERENTIAL EQUATION IS AN ORDINARY DIFFERENTIAL EQUATION (ODE) IF THE UNKNOWN FUNCTION DEPENDS ON ONLY ONE INDEPENDENT VARIABLE IF THE UNKNOWN FUNCTION DEPENDS ON TWO OR MORE INDEPENDENT VARIABLES, THE DIFFERENTIAL EQUATION IS A PARTIAL DIFFERENTIAL EQUATION (PDE) WITH THE EXCEPTIONS OF CHAPTERS 31 AND 34, THE PRIMARY FOCUS OF THIS BOOK WILL BE ORDINARY DIFFERENTIAL EQUATIONS EXAMPLE 1 2 EQUATIONS (1 1) THROUGH (1 4) ARE EXAMPLES, OF ORDINARY DIFFERENTIAL EQUATIONS, SINCE THE UNKNOWN FUNCTION Y DEPENDS SOLELY ON THE VARIABLE X EQUATION (1 5) IS A PARTIAL DIFFERENTIAL EQUATION, SINCE Y DEPENDS ON BOTH THE INDEPENDENT VARIABLES T AND X THE ORDER OF A DIFFERENTIAL EQUATION IS THE ORDER OF THE HIGHEST DERIVATIVE APPEARING IN THE EQUATION EXAMPLE 1 3 EQUATION (1 1) IS A FIRST-ORDER DIFFERENTIAL EQUATION; (1 2), (1 4), AND (1 5) ARE SECOND-ORDER DIFFERENTIAL EQUATIONS [NOTE IN (1 4) THAT THE ORDER OF THE HIGHEST DERIVATIVE APPEARING IN THE EQUATION IS TWO ] EQUATION (1 3) IS A THIRD-ORDER DIFFERENTIAL EQUATION CHAPTER IF THE INDEPENDENT VARIABLE IS X, BUT REPRESENTS D2Y/DP2 IF THE INDEPENDENT VARIABLE IS P OBSERVE THAT PARENTHESES ARE USED IN Y(N) TO DISTINGUISH IT FROM THE NTH POWER, YN IF THE INDEPENDENT VARIABLE IS TIME, USUALLY DENOTED BY T, PRIMES ARE OFTEN REPLACED BY DOTS THUS, ?Y, Ÿ, AND ???Y REPRESENT DY/DT, D2Y/DT2, AND D3Y/DT3, RESPECTIVELY SOLUTIONS A SOLUTION OF A DIFFERENTIAL EQUATION IN THE UNKNOWN FUNCTION Y AND THE INDEPENDENT VARIABLE X ON THE INTERVAL (, IS A FUNCTION Y(X) THAT SATISFIES THE DIFFERENTIAL EQUATION IDENTICALLY FOR ALL X IN ( EXAMPLE 1 4 IS Y(X) = C1 SIN 2X + C2 COS 2X, WHERE C1 AND C2 ARE ARBITRARY CONSTANTS, A SOLUTION OF Y?? + 4Y = 0? DIFFERENTIATING Y, WE FIND Y? = 2C1 COS 2X - 2C2 SIN 2X AND Y?? = - 4C1 SIN 2X - 4C2 COS 2X HENCE, Y?? + 4Y = (- 4C1 SIN 2X - 4C2 COS 2X) + 4(C1 SIN 2X + C2 COS 2X) = (- 4C1 + 4C1) SIN 2X + (- 4C2 + 4C2) COS 2X = THUS, Y = C1 SIN 2X + C2 COS 2X SATISFIES THE DIFFERENTIAL EQUATION FOR ALL VALUES OF X AND IS A SOLUTION ON THE INTERVAL (- `, `) EXAMPLE 1 5 DETERMINE WHETHER Y = X2 - 1 IS A SOLUTION OF (Y?)4 + Y2 = -1 NOTE THAT THE LEFT SIDE OF THE DIFFERENTIAL EQUATION MUST BE NONNEGATIVE FOR EVERY REAL FUNCTION Y(X) AND ANY X, SINCE IT IS THE SUM OF TERMS RAISED TO THE SECOND AND FOURTH POWERS, WHILE THE RIGHT SIDE OF THE EQUATION IS NEGATIVE SINCE NO FUNCTION Y(X) WILL SATISFY THIS EQUATION, THE GIVEN DIFFERENTIAL EQUATION HAS NO SOLUTION WE SEE THAT SOME DIFFERENTIAL EQUATIONS HAVE INFINITELY MANY SOLUTIONS (EXAMPLE 1 4), WHEREAS OTHER DIF FERENTIAL EQUATIONS HAVE NO SOLUTIONS (EXAMPLE 1 5) IT IS ALSO POSSIBLE THAT A DIFFERENTIAL EQUATION HAS EXACTLY ONE SOLUTION CONSIDER (Y?)4 + Y2 = 0, WHICH FOR REASONS IDENTICAL TO THOSE GIVEN IN EXAMPLE 1 5 HAS ONLY ONE SOLUTION Y ? 0 A PARTICULAR SOLUTION OF A DIFFERENTIAL EQUATION IS ANY ONE SOLUTION THE GENERAL SOLUTION OF A DIFFERENTIAL EQUATION IS THE SET OF ALL SOLUTIONS EXAMPLE 1 6 THE GENERAL SOLUTION TO THE DIFFERENTIAL EQUATION IN EXAMPLE 1 4 CAN BE SHOWN TO BE (SEE CHAPTERS 8 AND 9) Y = C1 SIN 2X + C2 COS 2X THAT IS, EVERY PARTICULAR SOLUTION OF THE DIFFERENTIAL EQUATION HAS THIS GENERAL FORM A FEW PARTICULAR SOLUTIONS ARE: (A) Y = 5 SIN 2X - 3 COS 2X (CHOOSE C1 = 5 AND C2 = - 3), (B) Y = SIN 2X (CHOOSE C1 = 1 AND C2 = 0), AND (C) Y ? (CHOOSE C1 = C2 = 0) THE GENERAL SOLUTION OF A DIFFERENTIAL EQUATION CANNOT ALWAYS BE EXPRESSED BY A SINGLE FORMULA AS AN EXAMPLE CONSIDER THE DIFFERENTIAL EQUATION Y? + Y2 = 0, WHICH HAS TWO PARTICULAR SOLUTIONS Y = 1/X AND Y ? 0 INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS A DIFFERENTIAL EQUATION ALONG WITH SUBSIDIARY CONDITIONS ON THE UNKNOWN FUNCTION AND ITS DERIVATIVES, ALL GIVEN AT THE SAME VALUE OF THE INDEPENDENT VARIABLE, CONSTITUTES AN INITIAL-VALUE PROBLEM THE SUBSIDIARY CONDI TIONS ARE INITIAL CONDITIONS IF THE SUBSIDIARY CONDITIONS ARE GIVEN AT MORE THAN ONE VALUE OF THE INDEPENDENT VARIABLE, THE PROBLEM IS A BOUNDARY-VALUE PROBLEM AND THE CONDITIONS ARE BOUNDARY CONDITIONS BASIC CONCEPTS [CHAP EXAMPLE 1 7 THE PROBLEM Y?? + 2Y? = EX; Y(P) = 1, Y?(P) = 2 IS AN INITIAL-VALUE PROBLEM, BECAUSE THE TWO SUBSIDIARY CON DITIONS ARE BOTH GIVEN AT X = P THE PROBLEM Y?? + 2Y? = EX; Y(0) = 1, Y(1) = 1 IS A BOUNDARY-VALUE PROBLEM, BECAUSE THE TWO SUBSIDIARY CONDITIONS ARE GIVEN AT THE DIFFERENT VALUES X = 0 AND X = 1 A SOLUTION TO AN INITIAL-VALUE OR BOUNDARY-VALUE PROBLEM IS A FUNCTION Y(X) THAT BOTH SOLVES THE DIFFERENTIAL EQUATION AND SATISFIES ALL GIVEN SUBSIDIARY CONDITIONS SOLVED PROBLEMS 1 1 DETERMINE THE ORDER, UNKNOWN FUNCTION, AND THE INDEPENDENT VARIABLE IN EACH OF THE FOLLOWING DIFFERENTIAL EQUATIONS: (A) Y??? - 5XY? = EX + (B) ?? ?? + = ?+ T (SIN ) TY T Y Y T T (C) S D T DS ST DT DS S 2 + = (D) D B DP DB DP B B P ? ? ? ? ? ?+ ? ?? ? ?? + ? = (A) THIRD-ORDER, BECAUSE THE HIGHEST-ORDER DERIVATIVE IS THE THIRD THE UNKNOWN FUNCTION IS Y; THE INDEPENDENT VARIABLE IS X (B) SECOND-ORDER, BECAUSE THE HIGHEST-ORDER DERIVATIVE IS THE SECOND THE UNKNOWN FUNCTION IS Y; THE INDEPENDENT VARIABLE IS T (C) SECOND-ORDER, BECAUSE THE HIGHEST-ORDER DERIVATIVE IS THE SECOND THE UNKNOWN FUNCTION IS T; THE INDEPENDENT VARIABLE IS S (D) FOURTH-ORDER, BECAUSE THE HIGHEST-ORDER DERIVATIVE IS THE FOURTH RAISING DERIVATIVES TO VARIOUS POWERS DOES NOT ALTER THE NUMBER OF DERIVATIVES INVOLVED THE UNKNOWN FUNCTION IS B; THE INDEPENDENT VARIABLE IS P 1 2 DETERMINE THE ORDER, UNKNOWN FUNCTION, AND THE INDEPENDENT VARIABLE IN EACH OF THE FOLLOWING DIFFERENTIAL EQUATIONS: (A) Y D X DY Y = + (B) Y DX DY X ? ?? ? ??= + (C) ??? ? + ? = X X X (D) 17Y(4) - T6Y(2) - 4 2Y5 = 3 COS T (A) SECOND-ORDER THE UNKNOWN FUNCTION IS X; THE INDEPENDENT VARIABLE IS Y (B) FIRST-ORDER, BECAUSE THE HIGHEST-ORDER DERIVATIVE IS THE FIRST EVEN THOUGH IT IS RAISED TO THE SECOND POWER THE UNKNOWN FUNCTION IS X; THE INDEPENDENT VARIABLE IS Y (C) THIRD-ORDER THE UNKNOWN FUNCTION IS X; THE INDEPENDENT VARIABLE IS T (D) FOURTH-ORDER THE UNKNOWN FUNCTION IS Y; THE INDEPENDENT VARIABLE IS T NOTE THE DIFFERENCE IN NOTATION BETWEEN THE FOURTH DERIVATIVE Y(4), WITH PARENTHESES, AND THE FIFTH POWER Y5, WITHOUT PARENTHESES 1 3 DETERMINE WHETHER Y(X) = 2E-X + XE-X IS A SOLUTION OF Y?? + 2Y? + Y = 0 DIFFERENTIATING Y(X), IT FOLLOWS THAT Y?(X) = - 2E-X + E-X - XE-X = - E-X - XE-X Y??(X) = E-X - E-X + XE-X = XE-X SUBSTITUTING THESE VALUES INTO THE DIFFERENTIAL EQUATION, WE OBTAIN Y?? + 2Y? + Y = XE-X + 2(- E-X – XE-X) + (2E-X + XE-X) = THUS, Y(X) IS A SOLUTION CHAP 1] BASIC CONCEPTS 1 4 IS Y(X) ? 1 A SOLUTION OF Y?? + 2Y? + Y = X? FROM Y(X) ? 1 IT FOLLOWS THAT Y?(X) ? 0 AND Y??(X) ? 0 SUBSTITUTING THESE VALUES INTO THE DIFFERENTIAL EQUATION, WE OBTAIN Y?? + 2Y? + Y = 0 + 2(0) + 1 = 1 ? X THUS, Y(X) ? 1 IS NOT A SOLUTION 1 5 SHOW THAT Y = LN X IS A SOLUTION OF XY?? + Y? = 0 ON ( = (0, `) BUT IS NOT A SOLUTION ON ( = (- `, `) ON (0, `) WE HAVE Y? = 1/X AND Y?? = - 1/X2 SUBSTITUTING THESE VALUES INTO THE DIFFERENTIAL EQUATION, WE OBTAIN XY Y X X X ?? + ? = ? ? ?? ? ??+ = THUS, Y = LN X IS A SOLUTION ON (0, `) NOTE THAT Y = LN X COULD NOT BE A SOLUTION ON (- `, `), SINCE THE LOGARITHM IS UNDEFINED FOR NEGATIVE NUMBERS AND ZERO 1 6 SHOW THAT Y = 1/(X2 - 1) IS A SOLUTION OF Y? + 2XY2 = 0 ON ( = (-1, 1) BUT NOT ON ANY LARGER INTERVAL CONTAINING ( ON (-1, 1), Y = 1/(X2 - 1) AND ITS DERIVATIVE Y? = - 2X/(X2 - 1)2 ARE WELL-DEFINED FUNCTIONS SUBSTITUTING THESE VALUES INTO THE DIFFERENTIAL EQUATION, WE HAVE ? + = ? ? + ? ? ?? ? ??= Y XY X X X X ( ) THUS, Y = 1/(X2 - 1) IS A SOLUTION ON ( = (-1, 1) NOTE, HOWEVER, THAT 1/(X2 - 1) IS NOT DEFINED AT X = ±1 AND THEREFORE COULD NOT BE A SOLUTION ON ANY INTERVAL CONTAINING EITHER OF THESE TWO POINTS 1 7 DETERMINE WHETHER ANY OF THE FUNCTIONS (A) Y1 = SIN 2X, (B) Y2(X) = X, OR (C) Y X X 1); HENCE IT IS NOT A SOLUTION TO THE INITIAL-VALUE PROBLEM (B) Y2(X) SATISFIES BOTH INITIAL CONDITIONS BUT DOES NOT SATISFY THE DIFFERENTIAL EQUATION; HENCE Y2(X) IS NOT A SOLUTION (C) Y3(X) SATISFIES THE DIFFERENTIAL EQUATION AND BOTH INITIAL CONDITIONS; THEREFORE, IT IS A SOLUTION TO THE INITIAL-VALUE PROBLEM 1 8 FIND THE SOLUTION TO THE INITIAL-VALUE PROBLEM Y? + Y = 0; Y(3) = 2, IF THE GENERAL SOLUTION TO THE DIFFERENTIAL EQUATION IS KNOWN TO BE (SEE CHAPTER 8) Y(X) = C1E-X, WHERE C1 IS AN ARBITRARY CONSTANT SINCE Y(X) IS A SOLUTION OF THE DIFFERENTIAL EQUATION FOR EVERY VALUE OF C1, WE SEEK THAT VALUE OF C1 WHICH WILL ALSO SATISFY THE INITIAL CONDITION NOTE THAT Y(3) = C1E-3 TO SATISFY THE INITIAL CONDITION Y(3) = 2, IT IS SUFFICIENT TO CHOOSE C1 SO THAT C1E-3 = 2, THAT IS, TO CHOOSE C1 = 2E3 SUBSTITUTING THIS VALUE FOR C1 INTO Y(X), WE OBTAIN Y(X) = 2E3E-X = 2E3-X AS THE SOLUTION OF THE INITIAL-VALUE PROBLEM 1 9 FIND A SOLUTION TO THE INITIAL-VALUE PROBLEM Y?? + 4Y = 0; Y(0) = 0, Y?(0) = 1, IF THE GENERAL SOLUTION TO THE DIFFERENTIAL EQUATION IS KNOWN TO BE (SEE CHAPTER 9) Y(X) = C1 SIN 2X + C2 COS 2X SINCE Y(X) IS A SOLUTION OF THE DIFFERENTIAL EQUATION FOR ALL VALUES OF C1 AND C2 (SEE EXAMPLE 1 4), WE SEEK THOSE VALUES OF C1 AND C2 THAT WILL ALSO SATISFY THE INITIAL CONDITIONS NOTE THAT Y(0) = C1 SIN 0 + C2 COS 0 = C2 TO SATISFY THE FIRST INITIAL CONDITION, Y(0) = 0, WE CHOOSE C2 = 0 FURTHERMORE, Y?(X) = 2C1 COS 2X - 2C2 SIN 2X; THUS, = SUBSTITUTING THESE VALUES OF C1 AND C2 INTO Y(X), WE OBTAIN Y X X ( ) SIN = 2 AS THE SOLUTION OF THE INITIAL-VALUE PROBLEM 1 10 FIND A SOLUTION TO THE BOUNDARY-VALUE PROBLEM Y?? + 4Y = 0; Y(P/8) = 0, Y(P/6) = 1, IF THE GENERAL SOLUTION TO THE DIFFERENTIAL EQUATION IS Y(X) = C1 SIN 2X + C2 COS 2X NOTE THAT Y C C C C ? ? ? ? ?? ? ??= ? ?? ? ??+ ? ?? ? ??= ? ?? ? ??+ ? ?? SIN COS ? ?? TO SATISFY THE CONDITION Y(P/8) = 0, WE REQUIRE C C ? ?? ? ??+ ? ?? ? ??= (1) FURTHERMORE, Y C C C C ? ? ? ? ?? ? ??= ? ?? ? ??+ ? ?? ? ??= ? ?? ? ??+ ? ?? ? SIN COS ?? TO SATISFY THE SECOND CONDITION, Y(P/6) = 1, WE REQUIRE = ? = ? C SUBSTITUTING THESE VALUES INTO Y(X), WE OBTAIN Y X X X ( ) (SIN COS ) = ? ? AS THE SOLUTION OF THE BOUNDARY-VALUE PROBLEM 1 11 FIND A SOLUTION TO THE BOUNDARY-VALUE PROBLEM Y?? + 4Y = 0; Y(0) = 1, Y(P/2) = 2, IF THE GENERAL SOLUTION TO THE DIFFERENTIAL EQUATION IS KNOWN TO BE Y(X) = C1 SIN 2X + C2 COS 2X SINCE Y(0) = C1 SIN 0 + C2 COS 0 = C2, WE MUST CHOOSE C2 = 1 TO SATISFY THE CONDITION Y(0) = 1 SINCE Y(P/2) = C1 SIN P + C2 COS P = - C2, WE MUST CHOOSE C2 = -2 TO SATISFY THE SECOND CONDITION, Y(P/2) = 2 THUS, TO SATISFY BOTH BOUNDARY CONDITIONS SIMULTANEOUSLY, WE MUST REQUIRE C2 TO EQUAL BOTH 1 AND - 2, WHICH IS IMPOSSIBLE THEREFORE, THERE DOES NOT EXIST A SOLUTION TO THIS PROBLEM 1 12 DETERMINE C1 AND C2 SO THAT Y(X) = C1 SIN 2X + C2 COS 2X + 1 WILL SATISFY THE CONDITIONS Y(P/8) = 0 AND ? = Y ( / ) ? NOTE THAT Y C C C C ? ? ? ? ?? ? ??= ? ?? ? ??+ ? ?? ? ??+ = ? ?? ? ??+ ? SIN COS ?? ? ??+ TO SATISFY THE CONDITION Y(P/8) = 0, WE REQUIRE C C ( ) ( ) , + + = OR EQUIVALENTLY, C C + = ? (1) CHAP 1] BASIC CONCEPTS SINCE Y?(X) = 2C1 COS 2X - 2C2 SIN 2X, ?? ?? ? ??= ? ?? ? ??? ? ?? ? ?? = ? ?? ? ??? Y C C C C ? ? ? COS SIN ? ?? ? ??= ? C C TO SATISFY THE CONDITION ? = Y ( / ) , ? 2 WE REQUIRE C C ? = , OR EQUIVALENTLY, C1 - C2 = = ? + ( ) 1 13 DETERMINE C1 AND C2 SO THAT Y(X) = C1E2X + C2EX + 2 SIN X WILL SATISFY THE CONDITIONS Y(0) = 0 AND Y?(0) = 1 BECAUSE SIN 0 = 0, Y(0) = C1 + C2 TO SATISFY THE CONDITION Y(0) = 0, WE REQUIRE C1 + C2 = (2) SOLVING (1) AND (2) SIMULTANEOUSLY, WE OBTAIN C1 = - AND C2 = 1 SUPPLEMENTARY PROBLEMS IN PROBLEMS 1 14 THROUGH 1 23, DETERMINE (A) THE ORDER, (B) THE UNKNOWN FUNCTION, AND (C) THE INDEPENDENT VARIABLE FOR EACH OF THE GIVEN DIFFERENTIAL EQUATIONS 1 14 (Y??)2 - 3YY? + XY = 1 15 X4Y(4) + XY??? = EX 1 16 ?? ? ? = ? T S TS T SIN 1 17 Y(4) + XY??? + X2Y?? - XY? + SIN Y = 1 18 D X DY Y N N = + 1 19 D R DY D R DY Y DR DY ? ? ? ? ? ?+ + = 1 20 D Y DX Y X ? ? ? ? ? ? + = / 1 21 D B DP P = 1 22 DB DP P ? ?? ? ??= 1 23 Y(6) + 2Y4Y(3) + 5Y8 = EX 1 24 WHICH OF THE FOLLOWING FUNCTIONS ARE SOLUTIONS OF THE DIFFERENTIAL EQUATION Y? - 5Y = 0? (A) Y = 5, (B) Y = 5X, (C) Y = X5, (D) Y = E5X, (E) Y = 2E5X, (F) Y = 5E2X 1 25 WHICH OF THE FOLLOWING FUNCTIONS ARE SOLUTIONS OF THE DIFFERENTIAL EQUATION Y? - 3Y = 6? (A) Y = - 2, (B) Y = 0, (C) Y = E3X - 2, (D) Y = E2X - 3, (E) Y = 4E3X 2 , (C) Y ET = 2 , (D) Y ET = ? 2 , (E) Y ET = ? ? 1 28 WHICH OF THE FOLLOWING FUNCTIONS ARE SOLUTIONS OF THE DIFFERENTIAL EQUATION SHOWN BELOW? DY DX Y X XY = + (A) Y = X, (B) Y = X8 - X4, (C) Y X X = ? 1 29 WHICH OF THE FOLLOWING FUNCTIONS ARE SOLUTIONS OF THE DIFFERENTIAL EQUATION Y?? - Y = 0? (A) Y = EX, (B) Y = SIN X, (C) Y = 4E-X, (D) Y = 0, (E) Y X = + 1 30 WHICH OF THE FOLLOWING FUNCTIONS ARE SOLUTIONS OF THE DIFFERENTIAL EQUATION Y?? - XY? + Y = 0? (A) Y = X2, (B) Y = X, (C) Y = 1 - X2, (D) Y = 2X2 - 2, (E) Y = 1 31 WHICH OF THE FOLLOWING FUNCTIONS ARE SOLUTIONS OF THE DIFFERENTIAL EQUATION ?? ? ? + = X X X E IN PROBLEMS 1 36 THROUGH 1 39, FIND C SO THAT Y(X) = C(1 - X2) SATISFIES THE GIVEN INITIAL CONDITION 1 36 Y(0) = 1 1 37 Y(1) = 0 1 38 Y(2) = 1 1 39 Y(1) = IN PROBLEMS 1 40 THROUGH 1 49, FIND C1 AND C2 SO THAT Y(X) = C1 SIN X + C2 COS X WILL SATISFY THE GIVEN CONDITIONS DETERMINE WHETHER THE GIVEN CONDITIONS ARE INITIAL CONDITIONS OR BOUNDARY CONDITIONS 1 40 Y(0) = 1, Y?(0) = 1 41 Y(0) = 2, Y?(0) = 1 42 Y Y ? ? ? ?? ? ??= ?? ?? ? ??= , 1 43 Y Y ( ) , = ? ?? ? ??= ? 1 44 ? = ?? ?? ? ??= Y Y ( ) , ? 1 45 Y(0) = 1, Y?(P) = 1 46 Y(0) = 1, Y(P) = 1 47 Y(0) = 0, Y?(0) = 1 48 Y Y ? ? ? ?? ? ??= ? ?? ? ??= , 1 49 Y Y ( ) , = ?? ?? ? ??= ? CHAP 1] BASIC CONCEPTS 1 51 Y(X) = C1X + C2 + X2 - 1; Y(1) = 1, Y?(1) = 1 52 Y(X) = C1EX + C2E2X + 3E3X; Y(0) = 0, Y?(0) = 1 53 Y(X) = C1 SIN X + C2 COS X + 1; Y(P) = 0, Y?(P) =