Integrating factors from the ground up
The use of an integrating factor is a powerful method for solving differential equations (DEs) or, at least, simplifying their solution. A large class of first-order DEs are readily amenable to solution this way, and this article describes, step-by-step and from first principles -- why that is so, and how to do it. Despite its power, the use of an integrating factor is a somewhat obscure procedure, which rests on some rather fiddly theory, and can be awkward to remember. The theory is usually explained in textbooks in a highly abstract way, assuming a level of background knowledge that readers often don't have. My goal in this article is to explain the subject step-by-step, using specific examples to illustrate the theory, and documenting the procedure in the form of an algorithm. I'll start by describing how a specific kind of first-order DE can be rewritten in such a way that one side is expressed as just the derivative of a product, so that only integration of the other side is needed to get a solution. I'll show how this analysis leads to a simple formula that will allow a certain class of first order DEs to be solved systematically. Although limited in scope, the method is conceptually straightforward, and does not require any use of partial differentiation. I then explain in more detail what is meant by an exact DE, and show how to solve an exact DE in the general case, by integration and the solution of a subsidiary (hopefully simpler) DE. Finally, I explain how to find an integrating factor that will make a first-order DE exact in the general case. This article assumes at least basic familiarity with differential equations and calculus. Comments and corrections are welcome; please send them to the usual place.Overview
Consider this rather ugly-looking, non-linear differential equation (DE): $$ y' - \frac{y \, tan x}{2} = \frac{x}{2y \, cos x} $$ On the face of it, it looks pretty difficult to solve -- it does to me, anyway. However, if we are fortunate enough to have a sudden flash of insight, it's almost trivially simple. The insight is to recognize that this DE can, with nothing more than high-school algebra -- be rearranged to look like this: $$ 2y \, y' \, cos x - y^2 sin x = x $$ If you're wondering why that's an improvement, that's certainly understandable. The trick, however, lies in spotting that $$ 2y \, y' \, cos x - y^2 sin x = \frac{d}{dx} y^2 \, cos x$$ To see that this is so, apply the product rule and the chain rule to the right-hand side, and you'll see that it gives the left-hand side. What this means is that the rather ugly DE we started with can be rewritten in a much simpler form: $$ \frac{d}{dx} y^2 \, cos x = x$$ If we integrate both sides we get $$ y^2 \, cos x = \int x dx = \frac{x^2}{2} + C $$ and this we can rearrange to give an expression for y, again using no more than high-school algebra. Solving this equation required a certain amount of inspiration. Much as I'd like to be the kind of person who is prone to such mathematical epiphanies, I'm not, and I suspect few people are -- I knew the answer because I started with the answer and worked back to the question. Since in real science and engineering problems we usually can't do that, the question arises whether there is some systematic way in which we can transform an arbitrary DE into one that can be expressed in the form $$ \frac{d}{dx} f(x,y) = g(x)$$ If the DE can be expressed in this form -- which is an example of what is known in the math trade as an exact equation -- then the DE reduces to an integration problem. There are two distinct problems tangled up here. The first is recognizing when a diferential equation is, in fact, exact -- it's often not obvious (I suspect the example I gave above was not obvious to most people). The second is handling a situation where the equation is not exact, and no amount of shuffling the terms about with algebra changes that. For many first-order DEs, both these problems can be solved by the use of an integrating factor. By multiplying all terms in the DE by the integrating factor -- which we have to determine first -- we transform the DE into an exact equation, that can be solved by integration. In practice, finding an integrating factor is often difficult. Moreover, having found one we frequently find that we've transformed an intractable DE into an equally intractable integration puzzle. Sometimes, however, we get lucky.An example from first principles
Let's consider how we can find and use an integrating factor for a simple, first-order differential equation: $$ y' + 3y = x $$ Note that this DE is in 'conventional form', where the coefficient of y' is 1, and there are no y terms on the right-hand side. It's certainly easier to apply the process if we start of this way. Because this DE is so straightforward, we can see at once that it is not exact, but it almost is; that is, it has a y' term and a y term on the left hand side, making it look as if it could be rearranged into the form $$ \frac{d}{dx} y \, u(x) = x $$ so that the left-hand side will fall out as a result of applying the product rule, giving us: $$ \frac{d}{dx} y \, u(x) = u y' + u'y $$ But it just doesn't quite work with the DE as it stands -- so far as I can tell, no amount of simple symbol-juggling will allow us to find a function u that will substitute d(uy)/dx for the LHS of the original DE. That is, there is no u that satisfies $$ uy' + u'y = y' + 3y $$ If uy'=y', then u=1; but then u' can't equal 3. Now let's assume that, although the DE is not exact at present, there is some function of x, or y, or both, by which we can multiply through, which will make it exact. Let's try a function of x for now; I will call that function g(x). So we can write the DE as: $$ g(x) y' + g(x) 3y = g(x) \, x $$ In order to make the left-hand side obtainable by the product rule we need the following equality to hold: $$ uy' + u'y = g(x) \, y' + g(x) \, 3y $$ Equating coefficients gives: $$ u=g(x), u'=3\,g(x) $$ Differentiating the first of these equations gives: $$ u'=g'(x), u'=3\,g(x) $$ and so $$ g'(x)=3\,g(x) $$ So it seems we have a way to find g(x), but it involves solving another differential equation. Have we really gained anything here? Well, it turns out that this particular DE is very easy to solve -- we can separate variables, or just see by inspection that $$ g(x) = e^{3x} $$ This is the solution because $$ \frac{d}{dx} e^{3x} = 3 e^{3x} $$ Readers who are still awake should have spotted that I've cheated a bit here: solving this DE for g(x) will have given an arbitrary constant of integration as part of the solution, but I have ignored it. It's safe to do that here, because all we're looking for is something to multiply every term in the DE. Putting the constant of integration in won't hurt, but it isn't necessary, because it will cancel out in later steps of the procedure. Remember that we can multiply every term in an equation by anything at all -- the goal here is to find something that simplifies the solution process, not necessarily a part of the solution itself. Having determined g(x) we can rewrite the original DE: $$ e^{3x} \, y' + e^{3x} \, 3y = e^{3x} \, x $$ Now we know that the left-hand side of this equation can be written in the form $$ \frac{d}{dx} y \, u(x) $$ because we derived the integration factor starting from that assumption. Had the assumption been incorrect, we would not have been able to get an integrating factor. In that case, we might have had to try a function of y instead, or some other approach (see later pages in this article for more details.) We also know, from an earlier result, that $$ u=g(x) $$ So our DE can now be written: $$ \frac{d}{dx} y \, e^{3x} = x \, e^3x $$ and we can just integrate both sides to get: $$ y \, e^{3x} = \int x \, e^{3x} $$ We're not quite done, because to carry out the integration we'll need to integrate by parts (or some other trick). I'm not going to write out all the steps -- it's just too tedious. In the end, we get: $$ y \, e^{3x} = x \, e^{3x} - \frac{e^{3x}}{9} + C $$ so $$ y = \frac{x}{3} + \frac{C}{e^{3x}} - \frac{1}{9} $$ The constant C will have to be determined from the initial conditions, in the usual way. You'll note that to get to this point we had two hurdles to jump over: solving a differential equation to get the integrating factor, and then integrating the right-hand side of the equation to get the solution. It turns out that, in practice, the first of these hurdles is not a high one for most first-order equations; but the second can be insurmountable. In practice, we don't want to carry out a detailed analysis every time we have to solve a first-order DE. So it's time to see how the process can be generalised to a method -- a formula, even -- for DEs whose form I described on this page.Deriving a formula to solve certain first-order DEs
In the previous section I showed, using a specific DE, how an integrating factor could be derived which allowed the DE to be expressed as an integral. This turned a problem in solving a DE into a problem in calculating an integral. In this section, I'll show how this procedure can be generalized into a method that will solve a large class of first-order differential equations -- anything that can be expressed in the following form: $$ y'(x) + p(x)\,y(x) = q(x) $$ As all functions on this page will be functions of x, for brevity I will usually drop the '(x)' from function names. This derivation is more abstract than the previous example, because we aren't working with specific functions. Nevertheless, the procedure is exactly the same. Rather than ending up with a specific integrating factor g, we'll have an expression to derive g from p(x). From there, we can express the solution of the differential equation as a simple formula. Of course, although the formula is easy to write down, it involves the calculation of two integrals, either of which might be rather difficult in practice. Although this derivation is relatively straightforward, I doubt that knowing the details will help you remember the final result. So if you think you might just need the result to use in an examination, and don't care how it is derived, skip right to the end of this page. As before, I'll use g [ =g(x) ] to represent the integrating factor. So, multiplying the DE by the (as yet unknown) integrating factor g: $$ g\,y' + g\, p\,y = g\,q $$ We want the left-hand side to reduce to the form $$ \frac{d}{dx} u\,y $$ where u is a function (presently unknown) of x. Expanding this derivative gives: $$ uy' + u'y $$ which we want to be equal to the left-hand side of the DE, including the integrating factor. So: $$ uy' + u'y = g\,y'+g\,p\,y $$ Equating coefficients gives: $$ uy' = g\,y',\, u'y = g\,p\,y $$ So $$ u = g,\, u' = g\,p $$ Differentiating the first of these equations gives $$ u'=g', u'=g\,p$$ and thus: $$ g' = g\, p $$ Solving this simple DE by separating variables: $$ \int \frac{1}{g} = \int p\,dx $$ $$ log(g) + C = \int p\,dx $$ $$ g = e^{\int p\,dx - C} = k e^{\int p\,dx } $$ Since we're going to multiply every term in the DE by this new function g, we can ignore the constant of integration k as it will eventually cancel out, and just write the integrating factor as:$$ g(x) = e^{\int p(x)\,dx} $$
Notice that this integrating factor depends only on the term p(x) in the original DE. We could now proceed to multiply all the terms in the DE by the integrating factor, and solve it as before. But we can go one step further, and reduce the whole solution process to a formula. We know that the left-hand side of the DE can be reduced to $$ \frac{d}{dx} y\,u = y\,g $$ Because if it could not, we could not have determined a value for g. So replacing the left-hand side: $$ \frac{d}{dx} y\,g = g\, q $$ Integrating both sides: $$ y\,g + C = \int g\, q\, dx $$ $$ y\,g = \int g\, q\, dx - C $$ Since the constant C is arbitrary, it doesn't matter what sign we give it; conventionally it is written as positive in this application. This leads to the final result -- a formula that gives the solution of the DE:If a differential equation can be expressed in the following form: $$ y'(x) + p(x)\,y(x) = q(x) $$ Then the solution of the equation is given by $$ y(x) = \frac{\int g(x)\, q(x)\, dx + C}{g(x)} $$ where $$ g(x) = e^{\int p(x)\, dx} $$ and the constant of integration in the last expression can be ignored.
Using the integrating factor formula to solve a DE directly
In the previous section I showed how to derive a formula for the solution of a DE of a particular class, based on an integrating factor. Just as a reminder:If a differential equation can be expressed in the following form: $$ y'(x) + p(x)\,y(x) = q(x) $$ Then the solution of the equation is given by $$ y(x) = \frac{\int g(x)\, q(x)\, dx + C}{g(x)} $$ where $$ g(x) = e^{\int p(x)\, dx} $$
Let's apply this method to the example I used earlier: $$ y' + 3y = x $$ In this example p=3 and q=x. So the integrating factor g is given by: $$ g = e^{\int p\, dx} = e^{\int 3\, dx} = e^{3x} $$ Remember that the constant of integration can be ignored, because it would cancel in the following step. The solution of the DE is $$ y = \frac {\int g\,q\, dx + C}{g} $$ $$ = \frac {\int x\,e^{3x}\, dx + C}{g} $$ Evaluating the integral using integration by parts: $$ y = \frac {x e^{3x} + \frac{e^{3x}}{9} + C}{g} $$ $$ = x e^{3x} + \frac{1}{9} + \frac{C}{e^{3x}} $$ This is exactly the same result we obtained two pages back, up to the sign of the arbitrary constant. Whether using this explicit formula for the solution is easier than carrying out an analysis of the DE and working from first principles in unclear; it probably is if you can remember the formula, and the DE happens to have the appropriate form.A more general approach to exact differential equations
In the previous pages I explained how to determine an integrating factor for differential equations that could be put into the form $$ y'(x) + p(x)\,y(x) = q(x) $$ The integrating factor was arrived at by assuming that the left-hand side of the DE could be viewed as the differential of a product $$ \frac{d}{dx} u(x)\,y $$ The we expanded the differential and equated terms, to get a (simple) DE that gave us the integrating factor. The problem is that not all first-order DEs can be put into the necessary form. So now it's time to consider a complementary approach, which will work on a larger class of differential equations. This method is more complex in that it requires at least an elementary understanding of partial derivatives. Instead of considering the left-hand side of a DE as product, we'll consider it as the derivative of a function of both x and y. Differentiating this using the total derivative formula will give us an expansion with which we can equate terms against the original DE. The method will only work directly if the DE is exact, that is, if it is already in the form where the left-hand side can be expressed as a single derivative. If it isn't, we'll need to introduce an integrating factor. But first, we need to examine in more detail how to solve an exact DE, in situations where the corresponding derivative is not obvious (or where we don't even know whether the DE is exact in the first place, which is usual). The first step in the process will be to write the differential equation in the form: $$ M(x,y) + N(x,y)\frac{dy}{dx} = 0 $$ And we'll end up rewriting this equation in the form $$ \frac{d}{dx} f(x,y(x)) = 0 $$ If we can do this, then the solution of the DE follows immediately, because by integrating we get $$ f(x,y(x)) = C $$
Note: The right-hand side of the DE does not have to be zero. It generally can be zero, because its terms can be incorporated into M or N. You can't leave any y terms on the right-hand side, because anything remaining there will eventually have to be integrated with respect to x. Sometimes it might be easier to leave some x terms on the right, and integrate at the end -- this might make it easier to find an integrating factor (at the expense of creating another integration challenge). I don't discuss this complexity further in this article, but it's something to bear in mind in practice. |
I evaluated both integrals in the example above, just to make the differences between them clear. In practice we only need one integral, and we can pick the easiest one. So, in this case, let's use the result derived from N that $$ f(x,y) = y^2 cos x + D(x)$$ Since we get N by differentiating f with respect to y, we should be able to get M by differentiating this value of f with respect to x. We have to differentiate D, of course, when we do this. $$ M(x,y) = \frac{d}{dx} (y^2 cos x + D(x)) = -y^2 sin x + D'(x) $$ But we know from the original DE that $$ M(x,y) = -y^2\,sin x - x $$ so $$ M(x,y) = -y^2\,sin x - x = -y^2 sin x + D'(x) $$ which leads to $$ D'(x) = -y^2 sin x + x + y^2 sin x = -x $$ Integrating D gives $$ D(x) = -\frac{x^2}{2} + E $$ So the value of f derived from integrating N(x,y) is $$ f(x,y) = y^2 cos x - \frac{x^2}{2} + E $$ E is a constant of integration and, this time, it actually is a constant.
We can heave a deep sigh of relief at this point, because the value of f arrived at here is the same (up to a constant) as the one derived by integrating M earlier. Had these turned out to be wildly different, then we would been sure of a mistake somewhere. The value of f we've calculated here is the solution to the differential equation: $$ y^2 cos x - \frac{x^2}{2} + E = c$$ But since E and c are both arbitrary constants, we can collapse them together: $$ y^2 cos x - \frac{x^2}{2} = C$$ This is the same result we arrived at back on the first page of this article. So let's sum up the steps in handling exact DEs, or DEs that might be suspected to be exact.
1. Write the DE in the following form:
$$ M(x,y) + N(x,y)\frac{dy}{dx} = 0 $$
2. Check that
$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$
If the equality holds, the DE is exact, and we can proceed.
3. Integrate N(x,y) with respect to x to get a
result for f(x,y), including a 'constant' of integration,
that is actually a function of x.
4. Partially differentiate f(x,y) with respect to x,
including the constant.
5. Set this derivative equal to M(x,y), and solve for the
'constant'.
6. Substitute the 'constant' back into the f(x,y) from step 3.
Set f(x,y)=c, where c is a constant. This is the solution to the
DE.
7. Rearrange the solution to get an explicit solution in terms of
the independent variable if necessary (and possible).
Finding an integrating factor in the general case
In the last section, I explained how to check whether a differential equation was exact -- capable of being expressed as a single derivative -- and to solve if it if it was. This is all very well if the DE happens to be exact. If it is not, we can try to make it exact. A couple of pages back I explained how to get an integrating factor for DEs whose form is $$ y'(x) + p(x)y = q(x) $$ and showed that the integrating factor was always $$ g(x)=e^{\int p(x)\, dx} $$ The method I will explain now is based on the analysis on the previous page, and will work (in principle) with any first-order DE that can be expressed in this form: $$ M(x,y) + N(x,y)y' = 0 $$ I say 'in principle' because, again, although the method will work, we may still be left trying to integrate something intractable. For illustration, I will use the following DE: $$ [3xy + 2y^2] + [(x^2 + 2xy)]y' = 0 $$ As before, the square brackets indicate the grouping of terms to form M(x,y) and N(x,y). So we have: $$ M(x,y) = 3x+2y^2 $$ $$ N(x,y) = x^2 + 2xy $$ You probably won't be surprised to read that I've picked this example to demonstrate the technique without pages and pages of algebra; we aren't always so fortunate. We can see that this DE is not exact, using the differentiation technique described on the previous page: to be exact we require $$ \frac{\partial}{\partial y} M(x,y) = \frac{\partial}{\partial x} N(x,y) $$ and, unfortunately, $$ 4y \ne 2x + 2y $$ As before, we proceed by assuming that there is an integrating factor that will make the DE exact. That is, there is some function of x, or y, or both, by which we can multiply the DE and make it exact. As before, I will call the integrating factor g, and I'll assume that it is a function of x. If this assumption fails, we'll need to try a function of y and, if that fails, a function of x and y. So far as I know, there is no obvious way to determine which of these kinds of integrating factor will work simply by looking at the DE -- we might have to try more than one. With the integrating factor, the DE will be $$ g(x)M(x,y) + g(x)N(x,y)y' = 0 $$ In this case: $$ g(x)[3xy + 2y^2] + g(x)[(x^2 + 2xy)]y' = 0 $$ This gives us new M and N terms, which I will call M2 and N2. For this DE, $$ M2 = g(x)(3xy + 2y^2); N2 = g(x)(x^2+2xy) $$ For this modified DE to be exact, we need $$ \frac{\partial}{\partial y} M2(x,y) = \frac{\partial}{\partial x} N2(x,y) $$ or, specifically, $$ \frac{\partial}{\partial y} g(x)(3xy + 2y^2) = \frac{\partial}{\partial x} g(x)(x^2+2xy) $$ Evaluating the derivatives gives us: $$ g(x)(3x+4y) = g'(x)(x^2+2xy) + g(x)(2x+2y) $$ Evaluating the right-hand integral required the chain rule, because g(x) is a function of x; for the left-hand side, g(x) is treated as a constant when taking the partial derivative with respect to y. Rearranging and collecting terms, we get $$ g(x)=x g'(x) $$ because I've picked the numbers carefully to make it come out this way. Whatever the numbers, using this method is going to leave us with a DE to solve; we hope it is easier to solve than the original DE. In this case, the solution is obvious -- we need something that gives 1 when differentiated by x, and the only thing that fits the bill is x itself. So $$ g(x)=x $$ In a more realistic example, we'll usually be able to solve by separating variables, and hoping we end up with a manageable integral. The result of all this manipulation is that we now have something that we know is an exact DE: $$ x(3xy + 2y^2) + (x^2 + 2xy)xy' = 0 $$ I gave the step-by-step procedure for solving this DE on the previous page, so I'll just outline the steps here. Integrate N with respect to y: $$ f(x,y) = \int (x^2 + 2xy)x dy = x^3y + x^2y^2 + D(x) $$ Differentiate f(x,y) with respect to x, and set the result equal to M(x,y): $$ 3x^2y + 2y^2x + D'(x) = x(3xy+2y^2) $$ Solve for D'(x): $$ D'(x) = 3x^2y + 2y^2x - 3x^2y - 2y^2x = 0 $$ Another 'lucky' result! Integrate to get D(x): $$ D(x) = \int 0 \, dy = C $$ Substitute D(x) back into f(x,y): $$ f(x,y) = x^3y + x^2y^2 + C $$ So the (implicit) solution to the DE is $$ f(x,y) = x^3y + x^2y^2 = c $$ To get an explicit solution for y(x), you'll need to use the quadratic formula, or something of that sort. In summary, the procedure for finding an integrating factor that will make a DE exact is as follows.
1. Write the DE in this form:
$$ M(x,y) + N(x,y)y' = 0 $$
2. See if it is already exact, by comparing the partial derivative
of M with respect to y with the partial derivative of
N with respect to x. If the derivatives are equal, the
DE is exact; no further work is required to make it exact, so solve
it using the method described earlier in this article.
3. Let
$$ M2=g(x)M(x,y); N2=g(x)N(x,y) $$
4. Determine g(x) by solving the (hopefully straightforward)
DE
$$ \frac{\partial}{\partial y} M2(x,y) = \frac{\partial}{\partial x} N2(x,y) $$
5. If no solution can be obtained, try replacing g(x) with
g(y), then g(x,y).
6. If an integrating factor is obtained, rewrite the DE in the form
$$ gM(x,y) + gN(x,y)y' = 0 $$
and solve it using the method described on page 5.
Summary
I've explained, starting from first principles, how an integrating factor can be obtained for a large class of first-order differential equations. The full method is lengthy and, frankly, rather error-prone with pencil-and-paper calculation, so it might be wise to approach the solution of a first-order DE in the following way.
1. Can it be solved using straightforward techniques such as
separation of variables or a simple substitution? There's no point
employing a complex method when a simple one will do.
2. see if the DE can be arranged into the form
$$ y'(x) + p(x)y(x) = q(x) $$
If it can, the derive an integrating factor from p(x) as
described earlier, multiply through, rearrange, and integrate.
3. Can the DE be put into one of the well-studied forms, like
Bernoulli's or Ricatti's equations? These well-known classes of DE have
straightforward, methodological solution procedures. An integrating factor
might also
work in these cases, but the textbook procedure is likely to be quicker.
4. Arrange the DE into the form
$$ M(x,y) + N(x,y)y' = 0 $$
5. Test whether it is already exact and, if so, solve it directly,
by rearranging and integrating.
6. If it is not exact, attempt to derive an integrating factor using
the method on the previous section.