DierentialEquationsINTEGRATINGFACTORMETHODGrahamSMcDonaldA Tutorial Module for learning to solve 1storder linear dierential equationsqTableofcontentsqBeginTutorialc2004 g.s.mcdonald@salford.ac.ukTableofcontents1. Theory2. Exercises3. Answers4. Standardintegrals5. Tipsonusingsolutions6. AlternativenotationFullworkedsolutionsSection1: Theory 31. TheoryConsideranordinarydierential equation(o.d.e.) thatwewishtosolve to nd out how the variabley depends on the variablex.If the equation is rstorder then the highest derivative involved isa rst derivative.If itisalsoalinearequationthenthismeansthateachtermcaninvolve y either as the derivativedydxOR through a single factor of y .Any such linear rst order o.d.e. can be re-arranged to give the fol-lowing standard form:dydx + P(x)y = Q(x)whereP(x)andQ(x)arefunctionsof x,andinsomecasesmaybeconstants.TocBackSection1: Theory 4A linear rst order o.d.e. can be solved using the integrating factormethod.After writing the equation in standard form, P(x) can be identied.One then multiplies the equation by the following integrating factor:IF= e

P(x)dxThis factor is dened so that the equation becomes equivalent to:ddx(IFy) = IFQ(x),whereby integrating both sides with respect tox, gives:IF y = _IFQ(x) dxFinally, divisionbytheintegratingfactor(IF)gives yexplicitlyinterms ofx, i.e. gives the solution to the equation.TocBackSection2: Exercises 52. ExercisesIn each case, derive the general solution. When a boundarycondition is also given, derive the particular solution.Click onExercise links for full worked solutions (there are 10exercises in total)Exercise1.dydx + y = x; y(0) = 2Exercise2.dydx + y = ex; y(0) = 1Exercise3.x dydx + 2y = 10x2; y(1) = 3qTheoryqAnswersqIntegralsqTipsqNotationTocBackSection2: Exercises 6Exercise4.xdydx y = x2; y(1) = 3Exercise5.x dydx 2y = x4sin xExercise6.xdydx 2y = x2Exercise7.dydx + y cot x = cosec xqTheoryqAnswersqIntegralsqTipsqNotationTocBackSection2: Exercises 7Exercise8.dydx + y cot x = cos xExercise9.(x21)dydx + 2xy = xExercise10.dydx= y tan x sec x; y(0) = 1qTheoryqAnswersqIntegralsqTipsqNotationTocBackSection3: Answers 83. Answers1. General solution is y = (x 1) + Cex, and particularsolution is y = (x 1) + 3ex,2. General solution is y = ex(x + C) , and particular solution isy = ex(x + 1) ,3. General solution is y =52x2+Cx2, and particular solution isy =12(5x2+1x2) ,4. General solution is y = x2+ Cx , and particular solution isy = x2+ 2x ,5. General solution is y = x3cos x + x2sin x + Cx2,6. General solution is y = x2lnx + C x2,TocBackSection3: Answers 97. General solution is y sin x = x + C ,8. General solution is 4y sin x + cos 2x = C ,9. General solution is (x21)y =x22+ C ,10. General solution is y cos x = C x , and particular solution isy cos x = 1 x .TocBackSection4: Standardintegrals 104. Standardintegralsf (x)_f(x)dx f (x)_f(x)dxxn xn+1n+1(n = 1) [g (x)]ng

(x)[g(x)]n+1n+1(n = 1)1xln |x|g

(x)g(x)ln |g (x)|exexax axln a(a > 0)sin x cos x sinh x cosh xcos x sin x cosh x sinh xtan x ln |cos x| tanh x ln cosh xcosec x lntanx2cosech x lntanhx2sec x ln |sec x + tan x| sech x 2 tan1exsec2x tan x sech2x tanh xcot x ln |sin x| coth x ln |sinh x|sin2xx2 sin 2x4sinh2xsinh 2x4x2cos2xx2 +sin 2x4cosh2xsinh 2x4+x2TocBackSection4: Standardintegrals 11f (x)_f (x) dx f (x)_f (x) dx1a2+x21a tan1xa1a2x212a lna+xax(0 a>0)1a2x2sin1xa1a2+x2lnx+a2+x2a(a > 0)(a < x < a)1x2a2lnx+x2a2a(x>a>0)a2x2a22_sin1_xa_ a2+x2a22_sinh1 _xa_ +xa2+x2a2_+xa2x2a2_x2a2a22_cosh1 _xa_ +xx2a2a2_TocBackSection5: Tipsonusingsolutions 125. Tipsonusingsolutionsq When looking at the THEORY, ANSWERS, INTEGRALS, TIPSor NOTATION pages, use the Back button (at the bottom of thepage) to return to the exercises.q Use the solutions intelligently. For example, they can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are correct.q Try to make less use of the full solutions as you work your waythrough the Tutorial.TocBackSection6: Alternativenotation 136. AlternativenotationThe linear rst order dierential equation:dydx + P(x) y = Q(x)has the integrating factor IF=e

P(x) dx.The integrating factor method is sometimes explained in terms ofsimpler forms of dierential equation. For example, when constantcoecientsa andb are involved, the equation may be written as:a dydx + b y = Q(x)In our standard form this is:dydx +ba y =Q(x)awith an integrating factor of:IF= e

badx= ebxaTocBackSolutionstoexercises 14FullworkedsolutionsExercise1.Compare with form:dydx + P(x)y = Q(x) (P, Q are functions ofx)Integrating factor: P(x) = 1.Integrating factor, IF = e

P(x)dx= e

dx= exMultiply equation by IF:exdydx + exy = exxi.e.ddx [exy] = exxTocBackSolutionstoexercises 15Integrate both sides with respect tox:exy = ex(x 1) + C{ Note:_udvdxdx = uv _vdudx dx i.e. integration by parts withu x,dvdx ex xex_exdx xexex= ex(x 1) }i.e. y = (x 1) + Cex.Particular solution withy(0) = 2:2 = (0 1) + Ce0= 1 + C i.e. C = 3 and y = (x 1) + 3ex.Return to Exercise 1TocBackSolutionstoexercises 16Exercise2.Integrating Factor: P(x) = 1 , IF = e

Pdx= e

dx= exMultiply equation:exdydx + exy = exexi.e.ddx [exy] = 1Integrate:exy = x + Ci.e. y = ex(x + C) .Particular solution:y = 1x = 0gives 1 = e0(0 + C)= 1.C i.e. C = 1and y = ex(x + 1) .Return to Exercise 2TocBackSolutionstoexercises 17Exercise3.Equation is linear, 1st order i.e.dydx + P(x)y = Q(x)i.e.dydx +2xy = 10x, where P(x) =2x, Q(x) = 10xIntegrating factor : IF = e

P(x)dx= e2

dxx= e2 lnx= elnx2= x2.Multiply equation: x2dydx + 2xy = 10x3i.e.ddx_x2 y= 10x3Integrate: x2y =52x4+ Ci.e. y =52x2+Cx2TocBackSolutionstoexercises 18Particular solution y(1) = 3 i.e. y(x) = 3 whenx = 1.i.e. 3=52 1 +C1i.e.62=52 + Ci.e. C =12 y=52x2+12x2=12_5x2+1x2_.Return to Exercise 3TocBackSolutionstoexercises 19Exercise4.Standard form:dydx _1x_y = xCompare withdydx + P(x)y = Q(x) , giving P(x) = 1xIntegrating Factor: IF = e

P(x)dx= e

dxx= elnx= eln(x1)=1x.Multiply equation:1xdydx 1x2y = 1i.e.ddx_1x y_= 1Integrate:1xy = x + Ci.e. y = x2+ Cx.TocBackSolutionstoexercises 20Particular solution withy(1) = 3:3 = 1 + Ci.e. C = 2Particular solution is y = x2+ 2x.Return to Exercise 4TocBackSolutionstoexercises 21Exercise5.Linear iny :dydx 2xy = x3sin xIntegrating factor: IF = e2

dxx= e2 lnx= elnx2=1x2Multiply equation:1x2dydx 2x3y = xsin xi.e.ddx_ 1x2y_= xsin xIntegrate:yx2= xcos x_ 1 (cos x)dx+C

[Note: integration by parts,_udvdxdx = uv _vdudxdx, u = x,dvdx= sin x]i.e.yx2= xcos x + sin x + Ci.e. y = x3cos x + x2sin x + Cx2.Return to Exercise 5TocBackSolutionstoexercises 22Exercise6.Standard form:dydx _2x_y = xIntegrating Factor: P(x) = 2xIF = e

Pdx= e2

dxx= e2 lnx= elnx2=1x2Multiply equation:1x2dydx 2x3y =1xi.e.ddx_ 1x2y_=1xTocBackSolutionstoexercises 23Integrate:1x2y =_dxxi.e.1x2y = lnx + Ci.e. y = x2lnx + Cx2.Return to Exercise 6TocBackSolutionstoexercises 24Exercise7.Of the form:dydx + P(x)y = Q(x) (i.e. linear, 1st order o.d.e.)where P(x) = cot x.Integrating factor : IF = e

P(x)dx= e

cos xsin xdx_ e

f

(x)f(x)dx_= eln(sin x)= sin xMultiply equation : sin x dydx + sin x_cos xsin x_y =sin xsin xi.e. sin x dydx + cos x y = 1i.e.ddx [sin x y] = 1Integrate: (sin x)y = x + C . Return to Exercise 7TocBackSolutionstoexercises 25Exercise8.Integrating factor: P(x) = cot x =cos xsin xIF = e

Pdx= e

cos xsin xdx= eln(sin x)= sin x_Note:cos xsin x f

(x)f(x)_Multiply equation:sin x dydx + sin x y cos xsin x= sin x cos xi.e.ddx [sin x y] = sin x cos xIntegrate: y sin x =_sin x cos x dxTocBackSolutionstoexercises 26{ Note:_sin xcos xdx _f(x)f

(x)dx _fdfdx dx_fdf=12f2+ C }i.e. y sin x =12 sin2x + C=12 12(1 cos 2x) + Ci.e. 4y sin x +cos 2x = C

_where C

= 4C + 1= constant_.Return to Exercise 8TocBackSolutionstoexercises 27Exercise9.Standard form:dydx +_2xx21_y =xx21Integrating factor: P(x) =2xx21IF = e

Pdx= e

2xx21dx= eln(x21)= x21Multiply equation: (x21)dydx+ 2xy = x(the original form of the equation was half-way there!)i.e.ddx_(x21)y = xIntegrate: (x21)y =12x2+ C .Return to Exercise 9TocBackSolutionstoexercises 28Exercise10.P(x) = tan xQ(x) = sec xIF = e

tan x dx= e

sin xcos xdx= e+

sin xcos xdx= eln(cos x)= cos xMultiply by IF: cos xdydx cos x sin xcos xy = cos x sec xi.e.ddx [cos x y] = 1 i.e. y cos x = x + C .y(0) = 1 i.e. y = 1 when x = 0 givescos(0) = 0 + C C = 1i.e. y cos x = x + 1 .Return to Exercise 10TocBack