Linear First-Order Equations

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1 5 Linear First-Orer Equations Linear first-orer ifferential equations make up another important class of ifferential equations that commonly arise in applications an are relatively easy to solve (in theory) As with the notion of separability, the iea of linearity for first-orer equations can be viewe as a simple generalization of the notion of irect integrability, an a relatively straightforwar (though, perhaps, not so intuitively obvious) metho will allow us to put any first-orer linear equation into a form that can be relatively easily integrate We will erive this metho in a short while (after, of course, escribing just what it means for a first-orer equation to be linear ) By the way, the criteria given here for a ifferential equation being linear will be extene later to higher-orer ifferential equations, an a rather extensive theory will be evelope to hanle linear ifferential equations of any orer That theory is not neee here; in fact, it woul be of very limite value An, to be honest, the basic technics we ll evelop in this chapter are only of limite use when it comes to solving higher-orer linear equations However, these basic technics involve an integrating factor, which is something we ll be able to generalize a little bit later (in chapter 7) to help solve much more general first-orer ifferential equations 51 Basic Notions Definitions A first-orer ifferential equation is sai to be linear if an only if it can be written as or, equivalently, as = f (x) p(x)y (51) + p(x)y = f (x) (52) where p(x) an f (x) are known functions of x only Equation (52) is normally consiere to be the stanar form for first-orer linear equations Note that the only appearance of y in a linear equation (other than in the erivative) is in a term where y alone is multiplie by some formula of x If there is any other functions of y appearing in the equation after you ve isolate the erivative, then the equation is not linear 103

2 104 Linear First-Orer Equations! Example 51: Consier the ifferential equation Solving for the erivative, we get x + 4y x 3 = 0 = x3 4y x = x 2 4 x y, which is with p(x) = 4 x = f (x) p(x)y an f (x) = x 2 So this first-orer ifferential equation is linear Aing 4 / x y to both sies, we then get the equation in stanar form, + 4 x y = x 2, On the other han is not linear because of the y x y2 = x 2 In testing whether a given first-orer ifferential equation is linear, it oes not matter whether you attempt to rewrite the equation as or as = f (x) p(x)y + p(x)y = f (x) If you can put it into either form, the equation is linear You may prefer the first, simply because it is a natural form to look for after solving the equation for the erivative However, because the secon form (the stanar form) is more suite for the methos normally use for solving these equations, more experience workers typically prefer that form! Example 52: Consier the equation x 2 + x 3 [y sin(x)) = 0 Diviing through by x 2 an oing a little multiplication an aition converts the equation to + xy = x sin(x), which is the stanar form for a linear equation So this ifferential equation is linear

3 Basic Notions 105 It is possible for a linear equation + p(x)y = f (x) to also be a type of equation we ve alrea stuie For example, if p(x) = 0 then the equation is = f (x), which is irectly integrable If, instea, f (x) = 0, the equation can be rewritten as = p(x)y, showing that it is separable In aition, you can easily verify that a linear equation is separable if f (x) is any constant multiple of p(x) If a linear equation is also irectly integrable or separable, then it can be solve using methos alrea iscusse Otherwise, a small trick turns out to be very useful Deriving the Trick for Solving Suppose we want to solve some first-orer linear equation + py = f (53) (for brevity, p = p(x) an f = f (x) ) To avoi triviality, let s assume p(x) is not always 0 Whether f (x) vanishes or not will not be relevant The small trick to solving equation (53) comes from the prouct rule for erivatives: If µ an y are two functions of x, then µ [µy = y + µ Rearranging the terms on the right sie, we get [µy = µ + µ y, an the right sie of this equation looks a little like the left sie of equation (53) To get a better match, let s multiply equation (53) by µ, µ + µpy = µf With luck, the left sie of this equation will match the right sie of the last equation for the prouct rule, an we will have This, of course, requires that [µy = µ + µ y = µ (54) + µpy = µf µ = µp

4 106 Linear First-Orer Equations Assuming this requirement is met, the equations in (54) hol Cutting out the mile of that (an recalling that f an µ are functions of x only), we see that the ifferential equation reuces to [µy = µ(x) f (x) (55) The avantage of having our ifferential equation in this form is that we can actually integrate both sies with respect to x, with the left sie being especially easy since it is just a erivative with respect to x The function µ is calle an integrating factor for the ifferential equation As note in the erivation, it must satisfy µ = µp (56) This is a simple separable ifferential equation for µ (remember, p = p(x) is a known function) Any nonzero solution to this can be use as an integrating factor (the zero solution, µ = 0, woul simplify matters too much!) Applying the approach we learne for separable ifferential equations, we ivie through by µ, integrate, an solve the resulting equation for µ : 1 µ µ = p(x) ln µ = p(x) µ = ±e p(x) Since we only nee one function µ(x) satisfying requirement (56), we can rop both the ± an any arbitrary constant arising from the integration of p(x) This leaves us with a relatively simple formula for our integrating factor; namely, µ(x) = e p(x) (57) where it is unerstoo that we can let the constant of integration be zero 52 Solving First-Orer Linear Equations As we just erive, the real trick to solving a first-orer linear equation is to reuce it to an easily integrate form via the use of an integrating factor Let me outline a proceure for actually carrying out the necessary steps To illustrate these steps, we will immeiately use them to fin the general solution to the equation from example 51, The Proceure: x + 4y = x 3 1 Get the equation into the stanar form for first-orer linear ifferential equations, + p(x)y = f (x)

5 Solving First-Orer Linear Equations 107 For our example, we just ivie through by x, obtaining + 4 x y = x 2 As note in example 51, this is the esire form with p(x) = 4 x an f (x) = x 2 2 Compute an integrating factor µ(x) = e p(x) Remember, since we only nee one integrating factor, we can let the constant of integration be zero here For our example, µ(x) = e p(x) = e 4 x = e 4 ln x Applying some basic ientities for the natural logarithm, we can rewrite this last expression in a much more convenient form: µ(x) = e 4 ln x = e ln x4 = x 4 = x 4 3a Multiply the ifferential equation (in stanar form) by the integrating factor, [ µ + p(x)y = f (x) µ + µpy = µf, b an observe that, via the prouct rule an choice of µ, the left sie can be written as the erivative of the prouct of µ an y, µ + µpy = µf, }{{} [µy c an then rewrite the ifferential equation as [µy = µf, For our example, µ = x 4 Multiplying our equation by this an proceeing through the three substeps above, yiels x 4[ + 4 x y = x 2 x 4 + 4x 3 y = x 6 }{{} [x4 y [x 4 y = x 6

6 108 Linear First-Orer Equations 4 Integrate with respect to x both sies of the last equation obtaine, [µy = µ(x) f (x) µy = µ(x) f (x) Don t forget the arbitrary constant here! Integrating the last equation in our example, [x 4 y = x 6 x 4 y = 1 7 x 7 + c 5 Finally, solve for y by iviing through by µ For our example, [ y = x x 7 + c = 1 7 x 3 + cx 4 It is possible to use the above proceure to erive an explicit formula for computing y from p(x) an f (x) Unfortunately, it is not a particularly simple formula, an those who attempt to memorize it typically make more mistakes than those who simply remember the above proceure So I won t tell you that formula, yet 1! Example 53: Consier e x = ex y, y(0) = 7 Subtracting 3e x y from both sies an then multiplying through by e x ifferential equation into the esire form, puts this linear So p(x) = 3, an our integrating factor is 3y = 20e x µ = µ(x) = e 3 = e 3x Multiplying the ifferential equation by µ an following the rest of the steps in our proceure gives us the following: e 3x[ 3y = 20e x e 3x 3e 3x y = 20e 4x }{{} [e 3x y 1 If you must see this formula, glance ahea to theorem 51 on page 114

7 Solving First-Orer Linear Equations 109 [e 3x y = 20e 4x [e 3x y = 20e 4x e 3x y = 5e 4x + c y = e 3x [ 5e 4x + c So the general solution to our ifferential equation is Thus, y(x) = 5e x + ce 3x Using this formula for y(x) with the initial conition gives us 7 = y(0) = 5e 0 + ce 3 0 = 5 + c c = = 12, an the solution to the given initial-value problem is y(x) = 5e x + 12e 3x Let us briefly get back to our requirement for µ = µ(x) being an integrating factor for That requirement was equation (56), + py = f µ = µp Now, in computing this µ, you will often get something like µ(x) = µ 0 (x) where µ 0 (x) is a relatively simple continuous function (eg, µ(x) = sin(x) ) Consequently, on any interval over which the graph of µ 0 (x) never crosses the X axis, µ 0 (x) = µ(x) or µ 0 (x) = µ(x) Either way, µ 0 = [±µ = ± µ = ±µp = µ 0 p So µ 0 also satisfies the requirement for being an integrating factor for the given ifferential equation This means that, if in computing µ you o get something like µ(x) = µ 0 (x) where µ 0 (x) is a relatively simple function, then you can ignore the absolute value brackets an just use µ 0 for your integrating factor

8 110 Linear First-Orer Equations! Example 54: Consier solving the linear ifferential equation + cot(x)y = x csc(x) This equation is alrea in the esire form In a case like this, it is often a goo iea to see what the equation looks like in terms of sines an cosines, So [ cos(x) + y = sin(x) x sin(x) To fin µ = e p, first observe that, ignoring the constant of integration, cos(x) p(x) = sin(x) = sin(x) = ln sin(x) sin(x) µ = µ(x) = e p(x) = e ln sin(x) = sin(x) As iscusse above, we can just rop the an use sin(x) for the integrating factor Doing so, an stepping through the rest of our proceure, we have [ sin(x) + cos(x) sin(x) y = x sin(x) sin(x) + cos(x)y = x }{{} [sin(x)y [sin(x)y = x sin(x)y = 1 2 x 2 + c 1 y = x2 + c 2 sin(x) 53 On Using Definite Integrals with Linear Equations Integration arises twice in our metho for solving + p(x)y = f (x) It first arises when we integrate p to get the integrating factor, µ(x) = e p(x) It then is neee again when we then integrate both sies of the corresponing equation [µy = µf

9 On Using Definite Integrals with Linear Equations 111 At either point, of course, we coul use efinite integrals instea of inefinite integrals Let s first look at what happens when we integrate both sies of the last equation using efinite integrals Remember, everything is a function of x, so this equation can be written a bit more explicitly as [µ(x)y(x) = µ(x) f (x) As before, to avoi having x represent two ifferent entities, we replace the x s with another variable, say, s, an rewrite our current ifferential equation as [µ(s)y(s) = µ(s) f (s) s Then we pick a convenient lower limit a for our integration an integrate each sie of the above with respect to s from s = a to s = x, But So equation (58) reuces to a a [µ(s)y(s) s = s s [µ(s)y(s) s = µ(s)y(s) x a a µ(s) f (s) s (58) = µ(x)y(x) µ(a)y(a) µ(x)y(x) µ(a)y(a) = a µ(s) f (s) s, Solving this for y(x) yiels y(x) = 1 [ µ(a)y(a) + µ(x) a µ(s) f (s) s (59) This is not a simple enough formula to be worth memorizing (especially since you still have to remember what µ is) Nonetheless, it is a formula worth knowing about for at least two goo reasons: 1 This formula can automatically take into account an initial value y( ) = y 0 All we have to o is to choose the lower limit a to be Then formula (59) tells us that the solution to + py = f with y() = y 0 is y = 1 [ µ( )y 0 + µ(x) µ(s) f (s) s (510) 2 Even if we cannot etermine a relatively nice formula for integral of µf (for a given choice of µ an f ), the value of the integral in formula (59) can, in practice, still be accurately compute for esire values of x using numerical integration routines foun in stanar computer math packages Inee, using any of these packages an formula (59), you coul probably program a computer to accurately compute y(x) for a number of values of x an use these values to prouce a very accurate graph of y

10 112 Linear First-Orer Equations! Example 55: Consier solving 2xy = 4 with y(0) = 3 The ifferential equation is clearly linear an in the esire form for the first step of our proceure Computing the integrating factor, we fin that, here, µ = e p(x) = e [ 2x = e x2 +c Choosing, as we may, c to be zero, we then get µ(x) = e x2 Plugging this into formula (59) (an choosing a = 0 since we have y(0) = 3 as the initial conition) yiels y(x) = 1 µ(x) = 1 e x2 [ µ(0)y(0) + [ e = e x2 [ e s2 s µ(s) f (s) s e s2 4 s This is the solution to our initial-value problem The integral, 0 e s2 s, was left unevaluate because no one has yet foun a nice formula for this integral At best, we can hie this integral by using the error function (see page 30), rewriting our formula for y as [ y(x) = e x π erf(x) Still, to fin the value of, say, y(4), we woul have to either numerically approximate the integral in 4 y(4) = e [ e s2 s or look up the value of the error function in y(4) = e 42 [ π erf(4) Either way, a ecent computer math package coul be helpful 0 As alrea note, we coul also use a efinite integral in etermining the integrating factor This means µ woul be given by ( ) µ(x) = exp p(s) s a

11 Integrability an the Existence an Uniqueness of Solutions 113 where a was any appropriate lower limit Naturally, if we ha an initial conition y( ) = y 0, it woul make sense to let a = This woul slightly simplify formula (510) to y = 1 [ y 0 + µ(s) f (s) s (511) µ(x) since (0 ) µ( ) = exp p(s) s = e 0 = 1 In practice, there is little to be gaine in using a efinite integral in the computation of µ unless there is not a reasonable formula for the integral of p Then we are pretty well force into using a efinite integral to compute µ(x) an to computing this integral numerically for each value of x of interest That, in turn, woul pretty well force us to compute y(x) for each x of interest by using numerical computation of formula (510) 54 Integrability an the Existence an Uniqueness of Solutions If you check, you will see that our erivation of the efinite integral formula y(x) = 1 [ y 0 + µ(x) as a solution to the initial-value problem µ(s) f (s) s with + p(x)y = f (x) with y() = y 0 ( ) µ(x) = exp p(s) s merely require that y be any solution to this problem, an that p an f be sufficiently integrable for the existence of the integrals involving them That is, every solution to this initialvalue problem is given by this one formula Conversely, as long as p an f are sufficiently integrable, you can use elementary calculus to ifferentiate the above efinite integral formula an verify that the y efine by this formula is a solution to the above initial-value problem (see problem 55) Thus, the above efinite integral formula gives us the one an only solution to the above initial-value problem, provie p an f are sufficiently integrable Just what is sufficiently integrable? Basically, we want the integrals p(s) s an µ(s) f (s) s to be well-efine, continuous functions of x in whatever interval of interest (α, β) we have (Note that this ensures ( ) µ(x) = exp p(s) s is never zero in this interval) Certainly, p an f will be sufficiently integrable if they are continuous on (α, β) But continuity is not necessary; p an f can have a few iscontinuities

12 114 Linear First-Orer Equations provie these iscontinuities are not too ba In particular, we can allow the same piecewiseefine functions consiere back in section 24 That (along with theorem 21 on page 35) gives us the following existence an uniqueness theorem for initial-value problems involving first-orer linear ifferential equations Theorem 51 (existence an uniqueness) Let p an f be functions that are continuous except for, at most, a finite number of finite-jump iscontinuities in an interval (α,β) Also let an y 0 be any two numbers with α < < β Then the initial-value problem + p(x)y = f (x) with y() = y 0 has exactly one solution over the interval (α,β), an that solution is given by y(x) = 1 [ ( ) y 0 + µ(s) f (s) s with µ(x) = exp p(s) s µ(x) Aitional Exercises 51 Determine whether each of the following ifferential equations is or is not linear, an, if it is linear, rewrite the equation in stanar form, + p(x)y = f (x) a x 2 + 3x 2 y = sin(x) b y 2 c e g i xy2 = x + 3x 2 y = sin(x) = 1 + (xy + 3y)2 = 1 + xy + 3y f = 4y + 8 e2x = 0 h + 4y = y3 j x = sin(x) y + cos( x 2) = 827y 52 Using the methos evelope in this chapter, fin the general solution to each of the following first-orer linear ifferential equations: a c + 2y = 6 b + 2y = 20e3x = 4y + 16x e x + 3y 10x 2 = 0 f x 2 2xy = x + 2xy = sin(x)

13 Aitional Exercises 115 g x = x + 3y h cos(x) i x + (5x + 2)y = 20 x j 2 x + sin(x) y = cos2 (x) + y = x 2xe 53 Fin the solution to each of the following initial-value problems using the methos evelope in this chapter: a b c 3y = 6 with y(0) = 5 3y = 6 with y(0) = 2 + 5y = e 3x with y(0) = y = 20x 2 with y(1) = 10 e x = y + x 2 cos(x) with y ( ) π = 0 2 f (1 + x 2 ) = x [ 3 + 3x 2 y with y(2) = 8 54 Express the answer to each of the following initial-value problems in terms of efinite integrals: a + 6xy = sin(x) with y(0) = 4 b x 2 + xy = x sin(x) with y(2) = 5 c x y = x 2 e x2 with y(3) = 8 55 Let (α,β) be an interval, an let an y 0 be any two numbers with α < < β Assume p an f are functions continuous at all but, at most, a finite number of points in (α, β), an that each of these iscontinuities is a finite-jump iscontinuity Define µ(x) an y(x) by ( ) µ(x) = exp p(s) s an y(x) = 1 [ y 0 + µ(x) µ(s) f (s) s a Compute the first erivatives of µ an y b Verify that y satisfies the initial conition y( ) = y 0 as well as the ifferential equation on (α,β) + p(x)y = f (x)

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