INTEGRATING FACTOR METHOD - Salford

1 Differential EquationsINTEGRATING FACTOR METHODG raham S McDonaldA Tutorial Module for learning to solve 1storder linear differential equationslTable of contentslBegin Tutorialc of on using notationFull worked solutionsSection 1: Theory31. TheoryConsider an ordinary differential equation ( ) that we wish tosolve to find out how the variableydepends on the the equation isfirst orderthen the highest derivative involved isa first it is also alinearequation then this means that each term caninvolveyeither as the derivativedydxOR through a single FACTOR such linear first order can be re-arranged to give the fol-lowing standard form:dydx+P(x)y=Q(x)whereP(x) andQ(x) are functions ofx, and in some cases may 1.

2 Theory4A linear first order can be solved using theintegrating writing the equation in standard form,P(x) can be then multiplies the equation by the following INTEGRATING FACTOR :IF=e P(x)dxThis FACTOR is defined so that the equation becomes equivalent to:ddx(IFy) = IFQ(x),whereby INTEGRATING both sides with respect tox, gives:IFy= IFQ(x)dxFinally, division by the INTEGRATING FACTOR (IF) givesyexplicitly interms ofx, gives the solution to the 2: Exercises52. ExercisesIn each case, derive the general solution. When a boundarycondition is also given, derive the particular onExerciselinks for full worked solutions (there are 10exercises in total)Exercise +y=x;y(0) = 2 Exercise +y=e x;y(0) = 1 Exercise + 2y= 10x2;y(1) = 3lTheorylAnswerslIntegralslTipslNotation TocJJIIJIBackSection 2: Exercises6 Exercise y=x2;y(1) = 3 Exercise 2y=x4sinxExercise 2y=x2 Exercise +ycotx= cosecxlTheorylAnswerslIntegralslTipslNot ationTocJJIIJIBackSection 2: Exercises7 Exercise +y cotx= cosxExercise 9.

3 (x2 1)dydx+ 2xy=xExercise secx;y(0) = 1lTheorylAnswerslIntegralslTipslNotation TocJJIIJIBackSection 3: Answers83. solution isy= (x 1) +Ce x, and particularsolution isy= (x 1) + 3e x, solution isy=e x(x+C) , and particular solution isy=e x(x+ 1) , solution isy=52x2+Cx2, and particular solution isy=12(5x2+1x2) , solution isy=x2+Cx, and particular solution isy=x2+ 2x, solution isy= x3cosx+x2sinx+Cx2, solution isy=x2lnx+C x2,TocJJIIJIBackSection 3: solution isysinx=x+C, solution is 4ysinx+ cos 2x=C, solution is (x2 1)y=x22+C, solution isycosx=C x, and particular solution isycosx= 1 4: Standard integrals104.

4 Standard integralsf(x) f(x)dxf(x) f(x)dxxnxn+1n+1(n6= 1)[g(x)]ng (x)[g(x)]n+1n+1(n6= 1)1xln|x|g (x)g(x)ln|g(x)|exexaxaxlna(a >0)sinx cosxsinhxcoshxcosxsinxcoshxsinhxtanx ln|cosx|tanhxln coshxcosecxln tanx2 cosechxln tanhx2 secxln|secx+ tanx|sechx2 tan 1exsec2xtanxsech2xtanhxcotxln|sinx|cothx ln|sinhx|sin2xx2 sin 2x4sinh2xsinh 2x4 x2cos2xx2+sin 2x4cosh2xsinh 2x4+x2 TocJJIIJIBackSection 4: Standard integrals11f(x) f(x)dxf(x) f(x)dx1a2+x21atan 1xa1a2 x212aln a+xa x (0<|x|<a)(a >0)1x2 a212aln x ax+a (|x|> a>0)1 a2 x2sin 1xa1 a2+x2ln x+ a2+x2a (a >0)( a < x < a)1 x2 a2ln x+ x2 a2a (x>a>0) a2 x2a22[sin 1(xa) a2+x2a22[sinh 1(xa)+x a2+x2a2]+x a2 x2a2] x2 a2a22[ cosh 1(xa)+x x2 a2a2]TocJJIIJIBackSection 5: Tips on using solutions125.

5 Tips on using solutionslWhen looking at the THEORY, ANSWERS, INTEGRALS, TIPSor NOTATION pages, use theBackbutton (at the bottom of thepage) to return to the the solutions intelligently. For example, they can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are to make less use of the full solutions as you work your waythrough the 6: Alternative notation136. Alternative notationThe linear first order differential equation:dydx+P(x)y=Q(x)has the INTEGRATING FACTOR IF=e P(x) INTEGRATING FACTOR METHOD is sometimes explained in terms ofsimpler forms of differential equation.

6 For example, when constantcoefficientsaandbare involved, the equation may be written as:adydx+b y=Q(x)In our standard form this is:dydx+bay=Q(x)awith an INTEGRATING FACTOR of:IF=e badx=ebxaTocJJIIJIBackSolutions to exercises14 Full worked solutionsExercise with form:dydx+P(x)y=Q(x) (P, Qare functions ofx) INTEGRATING FACTOR :P(x) = FACTOR , IF =e P(x)dx=e dx=exMultiply equation by IF:exdydx+exy= [exy] =exxTocJJIIJIBackSolutions to exercises15 Integrate both sides with respect tox:exy=ex(x 1) +C{Note: udvdxdx=uv integration by parts withu x,dvdx ex xex exdx xex ex=ex(x 1)} (x 1) +Ce solution withy(0) = 2:2 = (0 1) +Ce0= 1 + 3 andy= (x 1) + 3e to Exercise 1 TocJJIIJIBackSolutions to exercises16 Exercise FACTOR :P(x) = 1,IF =e Pdx=e dx=exMultiply equation:exdydx+exy=exe [exy] = 1 Integrate:exy=x+ x(x+C).

7 Particular solution:y= 1x= 0gives 1 =e0(0 +C)= 1andy=e x(x+ 1).Return to Exercise 2 TocJJIIJIBackSolutions to exercises17 Exercise is linear, 1st order +P(x)y=Q(x) +2xy= 10x,whereP(x) =2x, Q(x) = 10xIntegrating FACTOR : IF =e P(x)dx=e2 dxx=e2 lnx=elnx2= equation:x2dydx+ 2xy= [x2 y]= 10x3 Integrate:x2y=52x4+ +Cx2 TocJJIIJIBackSolutions to exercises18 Particular solutiony(1) = 3 (x) = 3 whenx= 3 =52 1 + + y=52x2+12x2=12(5x2+1x2).Return to Exercise 3 TocJJIIJIBackSolutions to exercises19 Exercise form:dydx (1x)y=xCompare withdydx+P(x)y=Q(x) , givingP(x) = 1xIntegrating FACTOR : IF =e P(x)dx=e dxx=e lnx=eln(x 1)= equation:1xdydx 1x2y= [1xy]= 1 Integrate:1xy=x+ +Cx.

8 TocJJIIJIBackSolutions to exercises20 Particular solution withy(1) = 3:3 = 1 + 2 Particular solution isy=x2+ 2x .Return to Exercise 4 TocJJIIJIBackSolutions to exercises21 Exercise iny:dydx 2xy=x3sinxIntegrating FACTOR : IF =e 2 dxx=e 2 lnx=elnx 2=1x2 Multiply equation:1x2dydx 2x3y= [1x2y]=xsinxIntegrate:yx2= xcosx 1 ( cosx)dx+C [Note: integration by parts, udvdxdx=uv vdudxdx, u=x,dvdx= sinx] xcosx+ sinx+ x3cosx+x2sinx+ to Exercise 5 TocJJIIJIBackSolutions to exercises22 Exercise form:dydx (2x)y=xIntegrating FACTOR :P(x) = 2xIF =e Pdx=e 2 dxx=e 2 lnx=elnx 2=1x2 Multiply equation.

9 1x2dydx 2x3y= [1x2y]=1xTocJJIIJIBackSolutions to exercises23 Integrate:1x2y= lnx+ + to Exercise 6 TocJJIIJIBackSolutions to exercises24 Exercise the form:dydx+P(x)y=Q(x) ( linear, 1st order )whereP(x) = FACTOR :IF =e P(x)dx=e cosxsinxdx{ e f (x)f(x)dx}=eln(sinx)= sinxMultiply equation: sinx dydx+ sinx(cosxsinx)y= sinx dydx+ cosx y= [sinx y] = 1 Integrate: (sinx)y=x+C .Return to Exercise 7 TocJJIIJIBackSolutions to exercises25 Exercise FACTOR :P(x) = cotx=cosxsinxIF =e Pdx=e cosxsinxdx=eln(sinx)= sinx{Note:cosxsinx f (x)f(x)}Multiply equation:sinx dydx+ sinx y cosxsinx= sinx [sinx y] = sinx cosxIntegrate:ysinx= sinx cosx dxTocJJIIJIBackSolutions to exercises26{Note: sinxcosx dx f(x)f (x)dx fdfdx dx fdf=12f2+C} +C=12 12(1 cos 2x) + 4ysinx+ cos 2x=C (whereC = 4C+ 1= constant).

10 Return to Exercise 8 TocJJIIJIBackSolutions to exercises27 Exercise form:dydx+(2xx2 1)y=xx2 1 INTEGRATING FACTOR :P(x) =2xx2 1IF =e Pdx=e 2xx2 1dx=eln(x2 1)=x2 1 Multiply equation: (x2 1)dydx+ 2x y=x(the original form of the equation was half-way there!) [(x2 1)y]=xIntegrate: (x2 1)y=12x2+C .Return to Exercise 9 TocJJIIJIBackSolutions to exercises28 Exercise (x) = tanxQ(x) = secxIF =e tanxdx=e sinxcosxdx=e+ sinxcosxdx=eln(cosx)= cosxMultiply by IF: cosxdydx cosx sinxcosxy= cosx [cosx y] = x+ (0) = 1 1 whenx= 0 givescos(0) = 0 +C C= x+ to Exercise 10 TocJJIIJIBack