Transcription of SECOND ORDER (inhomogeneous)
1 Differential EquationsSECOND ORDER ( inhomogeneous )Graham S McDonaldA Tutorial Module for learning to solve 2ndorder ( inhomogeneous ) differential equationslTable of contentslBegin Tutorialc of on using solutionsFull worked solutionsSection 1: Theory31 TheoryThis Tutorial deals with the solution of SECOND ORDER linear swith constant coefficients (a,bandc), of the form:ad2ydx2+bdydx+cy=f(x) ( )The first stepis to find the general solution of the homogeneous equa-tion [ as ( ), except thatf(x) = 0].
2 This gives us the comple-mentary function SECOND stepis to find a particular solutionyPSof the full equa-tion ( ). Assume thatyPSis a more general form off(x), havingundetermined coefficients, as shown in the following table:TocJJIIJIBackSection 1: Theory4f(x)Form ofyPSk(a constant)Clinear inxCx+Dquadratic inxCx2+Dx+EksinpxorkcospxCcospx+Dsinpxke pxCepxsum of the abovesum of the aboveproduct of the aboveproduct of the above(wherepis a constant)Note: If the suggested form ofyPSalready appears in thecomplementary function then multiply thissuggested form ofyPSinto ( ) yields values for the undetermined coef-ficients (C,D, etc).
3 Then,General solution of ( ) =yCF+ 2: Exercises52 ExercisesFind the general solution of the following equations. Where boundaryconditions are also given, derive the appropriate particular onExerciselinks for full worked solutions (there are 13 exer-cises in total)[Notation:y =d2ydx2, y =dydx]Exercise 2y 3y= 6 Exercise + 5y + 6y= 2xExercise 3.(a)y + 5y 9y=x2(b)y + 5y 9y= cos 2x(c)y + 5y 9y=e4x(d)y + 5y 9y=e 2x+ 2 xExercise 2y= sin 2xlTheorylAnswerslDerivativeslFindingyCF lTipsTocJJIIJIBackSection 2: Exercises6 Exercise y=exExercise +y 2y=e 2xExercise 2y +y=exExercise + 8y + 17y= 2e 3x;y(0) = 2 andy( 2)= 0 Exercise +y 12y= 4e2x;y(0) = 7 andy (0) = 0 Exercise + 3y + 2y= 10 cos 2x;y(0) = 1 andy (0) = 0 Exercise + 4y + 5y= 2e 2x;y(0) = 1 andy (0) = 2 Exercise 2+4dxd +3x=e 3.
4 X=12anddxd = 2 at = 0 Exercise 2+ 4dyd + 5y= 6 sin lTheorylAnswerslDerivativeslFindingyCFlT ipsTocJJIIJIBackSection 3: Answers73 x+Be3x 2 , 2x+Be 3x+x3 518, solutions arey=yCF+yPSwhereyCF=Aem1x+Bem2x(m1,2= 52 12 61)and (a)yPS= 19x2 1081x 68729(b)yPS= 13269cos 2x+10269sin 2x(c)yPS=127e4x(d)yPS= 115e 2x+19x 1381, + x+Be x sin 2x4+ 2, +Be x+12xex,TocJJIIJIBackSection 3: +Be 2x 13xe 2x, (A+Bx)ex+12x2ex, 4x(Acosx+Bsinx) +e 3x;y=e 4x(cosx e 2sinx) +e 3x, +Be 4x 23e2x;y=327e3x+6521e 4x 23e2x, 2x+Be x 12cos 2x+32sin 2x;y=32e 2x 12cos 2x+32sin 2x, 2x(Acosx+Bsinx) + 2e 2x;y=e 2x(2 cosx) , 3 +Be 12 e 3 ;x=12(1 )e 3 , 2 (Acos +Bsin ) 34(cos sin ).
5 TocJJIIJIBackSection 4: Standard derivatives94 Standard derivativesf(x)f (x)f(x)f (x)xnn xn 1[g(x)]nn[g(x)]n 1g (x)lnx1x(x >0)lng(x)1g(x)g (x) (g(x)>0)exexaxaxlna(a >0)sinxcosxsinhxcoshxcosx sinxcoshxsinhxtanxsec2xtanhxsech2xcosecx cosecxcotxcosechx cosechxcothxsecxsecxtanxsechx sechxtanhxcotx cosec2xcothx cosech2xTocJJIIJIBackSection 4: Standard derivatives10f(x)f (x)f(x)f (x)tan 1x11+x2tanh 1x11 x2( 1< x <1)cos 1x 1 1 x2( 1< x <1)cosh 1x1 x2 1(x >1)sin 1x1 1 x2( 1< x <1)sinh 1x1 x2+1wherecosecx= 1/sinx ,secx= 1/cosx ,cotx= 1/tanxcosechx= 1/sinhx ,sechx= 1/coshx ,cothx= 1/tanhxTocJJIIJIBackSection 5: FindingyCF115 FindingyCFOne considers the differential equation with RHS = 0.
6 Substituting atrial solution of the formy=Aemxyields an auxiliary equation :am2+bm+c= will have two roots (m1andm2).The general solutionyCF, when RHS = 0, is then constructed fromthe possible forms (y1andy2) of the trial solution. The auxiliaryequation may have:i) real different roots,m1andm2 yCF=y1+y2=Aem1x+Bem2xor ii) real equal roots,m1=m2 yCF=y1+xy2= (A+Bx)em1xor iii) complex roots,p iq yCF=y1+y2 epx(Acosqx+Bsinqx)TocJJIIJIBackSection 6: Tips on using solutions126 Tips on using solutionslWhen looking at the THEORY, ANSWERS, DERIVATIVES, FIND-INGyCFor TIPS pages, use theBackbutton (at the bottom of thepage) to return to the exerciseslUse the solutions intelligently.
7 For example, they can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are correctlTry to make less use of the full solutions as you work your waythrough the TutorialTocJJIIJIBackSolutions to exercises13 Full worked solutionsExercise 2y 3y= 6 Auxiliary equation ( ) from the homogeneous equationy 2y 3y= 0,ism2 2m 3 = (m 3)(m+ 1) = 0 3, m2= different roots : homogeneous equation has general solutiony=Aem1x+ x+Be3x(complementary function).
8 TocJJIIJIBackSolutions to exercises14f(x) = 6suggests form of particular solutionyPS=C(C= undetermined constant)y PS=y PS= 0 whenyPS= constantsubstituteyPS=C: 0 + 0 3C= 6 solution,y=yCF+yPS=Ae x+Be3x to Exercise 1 TocJJIIJIBackSolutions to exercises15 Exercise + 5y + 6y= ism2+ 5m+ 6 = (m+ 2)(m+ 3) = 2, m2= 3,yCF=Ae 2x+Be (x) = 2xsuggests substitution ofyPS=Cx+D(C, Dare undeter-mined constants)y PS=C, y PS= 0, substitution gives0 + 5C+ 6(Cx+D) = 5C+ 6D+ 6Cx= 2xConstant term :5C+ 6D= 0 Coefficient ofx.
9 6C= 5C6= 518 TocJJIIJIBackSolutions to exercises16 General solution,y=yCF+ 2x+Be 3x+13x to Exercise 2 TocJJIIJIBackSolutions to exercises17 Exercise (a)-(d) have same homogeneous + 5y 9y= 0 with + 5m 9 = ( 5 25 ( 36)) 52 12 61 Real different roots:yCF=Aem1x+Bem2x,wherem1= 52+12 61m2= 52 12 61 .(a)y + 5y 9y=x2 TryyPS=Cx2+Dx+E,dyPSdx= 2Cx+D ,d2yPSdx2= 2 CTocJJIIJIBackSolutions to exercises18 Substitution: 2C+ 5(2Cx+D) 9(Cx2+Dx+E) = 2C+ 5D 9E= 0 (constant term)10Cx 9Dx= 0 (terms inx) 9Cx2=x2(terms inx2) 9C= 1 1910C 9D= 0 givesD=10C9= 10812C+ 5D 9E= 0 givesE=19(2C+ 5D)=19( 29 5081)=19( 6881)= 68729 General solution is:y=yCF+yPS=Aem1x+Bem2x 19x2 1081x to exercises19(b)y + 5y 9y= cos 2xTryyPS=Ccos 2x+Dsin 2xy PS= 2 Csin 2x+ 2 Dcos 2xy PS= 4 Ccos 2x 4 Dsin 2xSubstitute.
10 4 Ccos 2x 4 Dsin 2x+ 5 ( 2 Csin 2x+ 2 Dcos 2x) 9(Ccos 2x+Dsin 2x) = cos ( 4C+10D 9C) cos 2x (4D+10C+9D) sin 2x= cos ( 13C+ 10D) cos 2x (13D+ 10C) sin 2x= cos 2xCoefficients of cos 2x: 13C+ 10D= 1 (i)Coefficients of sin 2x:13D+ 10C= 0 (ii)TocJJIIJIBackSolutions to exercises20(ii) givesD= 10C13then (i) gives 13C 100C13= 13 13 10013= 13269=CthenD= 1013 ( 13269)= + 13269cos 2x+10269sin 2xGeneral solution is:y=yCF+yPS=Aem1x+Bem2x+1269(10 sin 2x 13 cos 2x).TocJJIIJIBackSolutions to exercises21(c)y + 5y 9y=e4xTryyPS= PS= 4Ce4xy PS= 16Ce4xSubstitute: 16Ce4x+ 20Ce4x 9Ce4x= (16C+ 20C 9C)e4x= (by comparing coefficients ofe4x) solution is:y=yCF+yPS=Aem1x+Bem2x+ to exercises22(d)y + 5y 9y=e 2x+ 2 xTryyPS=Ce 2x+Dx+E( sum of the forms kepx and linear inx ) PS= 2Ce 2x+Dy PS= +4Ce 2xSubstitute: 4Ce 2x 10Ce 2x+5D 9Ce 2x 9Dx 9E=e 2x+2 2x: 4C 10C 9C= 1 C= : 9D= 1 D=.
