Transcription of Integration and Differential Equations
1 2 Integration and Differential EquationsOften, when attempting to solve a Differential equation, weare naturally led to computing one ormore integrals after all, Integration is the inverse of differentiation. Indeed, we have alreadysolved one simple second-order Differential equation by repeated Integration (the one arising inthe simplest falling object model , starting on page 10). Letus now briefly consider the generalcase where Integration is immediately applicable, and alsoconsider some practical aspects ofusing both the indefinite integral and the definite Directly-Integrable EquationsWe will say that a given first-order Differential equation isdirectly integrableif (and only if) itcan be (re)written asdydx=f(x)( )wheref(x)is some known function of justx(noy s ).
2 More generally, anyNth-order differ-ential equation will be said to bedirectly integrableif and only if it can be (re)written asdNydxN=f(x)( )where, again,f(x)is some known function of justx(noy s or derivatives ofy).! Example :Consider the equationx2dydx 4x=6.( )Solving this equation for the derivative:x2dydx=4x+6 dydx=4x+ the right-hand side of the last equation depends only onx, we do havedydx=f(x) withf(x)=4x+6x2 .2324 Integration and Differential EquationsSo equation ( ) is directly integrable.! Example :Consider the equationx2dydx 4xy=6.( )Solving this equation for the derivative:x2dydx=4xy+6 dydx=4xy+ , the right-hand side of the last equation depends on bothxandy, not justx.
3 Soequation ( ) isnotdirectly a directly-integrable equation is easy: First solve for the derivative to get the equationinto form ( ) or ( ), then integrate both sides as many times as needed to eliminate thederivatives, and, finally, do whatever simplification seemsappropriate.! Example :Again, considerx2dydx 4x=6.( )In example , we saw that it is directly integrable and can be rewritten asdydx=4x+ both sides of this equation with respect tox(and doing a little algebra):Zdydxdx=Z4x+6x2dx( ) y(x)+c1=Z 4x+6x2 dx=4Zx 1dx+6Zx 2dx=4 ln|x|+c2 6x 1+c3wherec1,c2, andc3are arbitrary constants. Rearranging things slightly and lettingc=c2+c3 c1, this last equation simplifies toy(x)=4 ln|x| 6x 1+c.
4 ( )This is our general solution to Differential equation ( ). Since bothln|x|andx 1arediscontinuous atx=0, the solution can be valid over any interval not containingx= Exercise :Consider the Differential equation in example and explain why they,which is an unknown function ofx, makes it impossible to completely integrate both sides ofdydx=4xy+6x2with respect Using Indefinite On Using Indefinite IntegralsThis is a good point to observe that, whenever we take the indefinite integrals of both sides ofan equation, we obtain a bunch of arbitrary constants c1,c2,..(one constant for eachintegral) that can be combined into a single arbitrary constantc.
5 In the future, rather thannote all the arbitrary constants that arise and how they combine into a single arbitrary constantcthat is added to the right-hand side in the end, let us agree tosimply add thatcat the end. Let snot explicitly note all the intermediate arbitrary constants. If, for example, we had agreed to thisbefore doing the last example, then we could have replaced all that material from equation ( )to equation ( ) withZdydxdx=Z4x+6x2dx y(x)=Z 4x+6x2 dx=4Zx 1dx+6Zx 2dx=4 ln|x| 6x 1+ should simplify our computations a convention of implicitly combining all the arbitraryconstants also allows us to writey(x)=Zdydxdx( )instead ofy(x)+some arbitrary constant= our new convention, that some arbitrary constant is still in equation ( )
6 It s just beenmoved to the right-hand side of the equation and combined with the constants arising from theintegral , like you, this author will get tired of repeatedly saying wherecis an arbitraryconstant when it is obvious that thec(or thec1or theAor ..) that just appeared in theprevious line is, indeed, some arbitrary constant. So let usnot feel compelled to constantlyrepeat the obvious, and agree that, when a new symbol suddenly appears in the computation ofan indefinite integral, then, yes, that is an arbitrary constant. Remember, though, to use differentsymbols for the different constants that arise when integrating a function already involving anarbitrary constant.
7 ! Example :Consider solvingd2ydx2=18x2.( )Clearly, this is directly integrable and will require two integrations. The first Integration yieldsdydx=Zd2ydx2dx=Z18x2dx=183x3+ and Differential EquationsCutting out the middle leavesdydx=6x3+ this, we havey(x)=Zdydxdx=Z 6x3+c1 dx=64x4+c1x+ the general solution to equation ( ) isy(x)=32x4+c1x+ practice, rather than use the same letter with different subscripts for different arbitraryconstants (as we did in the above example), you might just want to use different letters, say,writingy(x)=32x4+ax+binstead ofy(x)=32x4+c1x+ sometimes prevents dumb mistakes due to bad On Using Definite IntegralsBasic IdeasWe have been using theindefiniteintegral to recovery(x)
8 Fromdy/dxvia the relationZdydxdx=y(x)+ ,cis some constant (which we ve agreed to automatically combine with other constantsfrom other integrals).We could just about as easily have used the correspondingdefiniteintegral relationZxadydsds=y(x) y(a)( )to recovery(x)from its derivative. Note that, here, we ve usedsinstead ofxto denote thevariable of Integration . This prevents the confusion that can arise when using the same symbolfor both the variable of integrationandthe upper limit in the integral. The lower limit,a, can bechosen to be any convenient value. In particular, if we are also dealing with initial values, thenit makes sense to setaequal to the point at which the initial values are given.
9 Thatway (as wewill soon see) we will obtain a general solution in which the undetermined constant is simplythe initial from getting it into the formdydx=f(x),there are two simple steps that should be taken before using the definite integral to solve afirst-order, directly-integrable Differential equation:On Using Definite a convenient value for the lower limit of integrationa. In particular, if the value ofy(x0)is given for some pointx0, seta= the Differential equation withsdenoting the variable instead ofx( , replacexwiths),dyds=f(s).( )After that, simply integrate both sides of equation ( ) with respect tosfromatox:Zxadydsds=Zxaf(s)ds y(x) y(a)=Zxaf(s) solve fory(x)by addingy(a)to both sides,y(x)=Zxaf(s)ds+y(a).
10 ( )This is a general solution to the given Differential equation. It should be noted that the integralhere is a definite integral. Its evaluation does not lead to any arbitrary constants. However, thevalue ofy(a), until specified, can be anything; soy(a)is the arbitrary constant in this generalsolution.! Example :Consider solving the initial-value problemdydx=3x2withy(2)= we know the value ofy(2), we will use2as the lower limit for our integrals. Rewritingthe Differential equation withsreplacingxgivesdyds= this with respect tosfrom2tox:Zx2dydsds=Zx23s2ds y(x) y(2)=s3 x2=x3 fory(x)(and computing23) then gives usy(x)=x3 8+y(2).
