Multiplying through by an integrating factor to make the left side a single derivative

0:00Two methods so far. One wanted the right hand side to come apart into a piece in x times a piece in y. The other wanted it unchanged when both variables are scaled together. Here is a third shape, and it is the easiest of the three to recognise. The derivative of y, plus something times y, equals something else. Both of those somethings must be functions of x alone. Call them P and Q.

0:31That is the whole shape. A derivative, a multiple of the unknown, and on the right a function of x with no y in it anywhere. It looks like a small class of equations. It is not. Almost every equation in this topic is one of these, and the method that solves them is a single trick applied once. This class has a name. These are called linear equations. And that name is worth stopping on, because it is the sort of name that gets used constantly and explained never.

1:09It does not mean the solutions are straight lines. They almost never are. You will see exponentials, sines, and reciprocals of squares before this video is over, and every one of them is the solution of a linear equation. What linear records is a property of the equation, not of its answer. It is this. The unknown and its derivative each appear to the first power, and neither of them appears inside anything else.

1:41Not squared. Not multiplied by each other. Not inside a sine, a logarithm or a square root. Just each of them, once, with a coefficient in front. And that property can be tested, rather than eyeballed. Here is the test. Take the left hand side and treat it as a machine: feed it a function and it hands you back the derivative plus P times the function. Now feed it a combination. Take one curve, multiply it by a number, add a second curve multiplied by another number, and put the whole thing in.

2:21If the machine is linear, what comes out is the same combination of what the two curves gave separately. The numbers pass straight through it. That is the whole content of the word. A linear operator carries a combination to the combination of the readings. It has been checked here at twenty readings, four pairs of numbers at five places, and it holds at every one. And the test earns its keep by refusing things.

2:53Take a left hand side with the unknown squared in it: the derivative plus y times y. Feed it a combination and the squaring cross multiplies the two curves. The reading is not the combination of the readings, and the test says so at the first place it looks. Take one with the unknown inside a sine. Same verdict, for the same reason. Neither of those is linear, and neither of them can be solved by anything in this video.

3:26So this is not a shape you learn to spot. It is a property you can decide. Three equations that pass, two that do not, and the test told them apart without anyone recognising anything. Now the method, and the whole of it fits in one sentence. Look hard at the left hand side. A derivative, plus a multiple of the thing being differentiated. You have seen that shape before, from the other direction.

3:57That is exactly what the product rule produces. Differentiate y times some function of x, and you get the derivative of y times that function, plus y times the derivative of the function. Two terms. A derivative and a multiple of the unknown. So the left hand side of a linear equation looks like the wreckage of a product rule. The method is to put it back together. Here is the fold, done slowly.

4:31The left side is the derivative of y plus P times y. Two terms. It is not yet the derivative of anything, because the coefficient on the first term is one and the coefficient on the second is P, and the product rule does not leave you a mismatched pair like that. So multiply the whole equation through by a function of x that has not been chosen yet. Call it M.

5:00Now the left side is M times the derivative of y, plus M P times y. And the question is whether M can be chosen so that those two terms are exactly what you get by differentiating M times y. Differentiate M times y and see what the product rule actually gives. It gives M times the derivative of y, plus y times the derivative of M. Set that against what we have. M times the derivative of y, plus M P times y.

5:37The first terms are identical. They cancel against each other and take no part in what follows. What is left is a single demand: y times the derivative of M must equal M P times y. The y cancels too. And the condition that survives is as small as a condition gets. The derivative of M equals P times M. That is an equation for M, and it is one we can already solve. It separates.

6:11The derivative of M over M equals P. The left side is the derivative of the logarithm of M. So integrate both sides. The logarithm of M equals the integral of P with respect to x. Exponentiate, and M is the exponential of the integral of P. That function has a name. It is the integrating factor, and it is written I dot F for short. Notice what it depends on. Only P. The right hand side of the original equation never entered the derivation and has no say in what you multiply by.

6:54There is something quietly missing from that derivation, and if you have been drilled properly it will bother you. We integrated P and wrote no constant. Every integration you have done up to now carried a constant. This one did not, and nothing was said about why. The reason is that it does not survive. A different constant in the exponent multiplies M by a fixed number, and that number divides straight out of the final line.

7:26This has been measured rather than asserted. The factor for one of the worked equations was multiplied by five different numbers, including a negative one, and every single one of them still gave an answer that solves the equation at every place tested. So dropping it there is free. Free there. Not free at the end. And it is worth being precise about the difference, because it is the same letter in both places and only one of the two omissions is harmless.

8:02When you drop the constant while finding M, you throw away a number that was going to cancel. When you drop it at the end of the solution, you throw away every curve in the family but one. Here is what that costs. Take the answer to one of the worked equations and offer it a data point. Among five different constants, exactly one gives a curve through that point.

8:28The answer with the letter dropped is the constant nought, and it is not the one the point picks. It misses. So here is the method in three steps, and it never gets longer than this. One. Put the equation in the shape. A bare derivative, coefficient one, plus P times y, equals Q. If the derivative arrives with something in front of it, divide first. This is not optional and it is where most of the marks are lost.

9:02Two. Compute the factor. M is the exponential of the integral of P. Simplify it. Three. Write down the answer in this form. y times M equals the integral of Q times M, plus a constant. That third line is the one to memorise. Not the version solved for y. This one. Everything after that is integration, and integration is a skill you already have. First worked example. The derivative of y minus y equals the cosine of x.

9:40Step one is already done: the derivative stands alone, P is minus one, Q is the cosine. Step two. The integral of minus one is minus x, so the factor is the exponential of minus x. Step three. y times the exponential of minus x equals the integral of the exponential of minus x times the cosine of x. And that integral is the interesting part. Integrate by parts and you get another integral. Integrate by parts again and the original integral comes back, with a minus sign in front of it.

10:23That is not a failure. Call the integral I, and you now have an equation in I. Solve it like any other unknown. Doing that gives the integral as half the exponential of minus x times the difference of the sine and the cosine. So y times the exponential of minus x equals that, plus a constant. Divide by the factor, and the answer is half the sine minus half the cosine, plus a constant times the exponential of x.

10:58Look hard at the last term. The constant is multiplied by the exponential of x. It did not stay behind when we divided. Dividing by the factor divided the constant too, and dividing by the exponential of minus x is multiplying by the exponential of x. Remember that term. It settles an argument later in this video. Second one. x times the derivative of y, plus twice y, equals x squared.

11:33And this one opens with the step people skip. The derivative has an x in front of it. The shape demands a bare derivative. Divide the whole equation by x. Now the derivative stands alone, P is twice the reciprocal of x, and Q is x. If you had read the coefficients off the original equation you would have taken P to be two, and everything after that would have been wrong.

12:02Step two. The integral of twice the reciprocal of x is twice the logarithm of x. So the factor is the exponential of twice a logarithm. And a factor in that state is unusable. You cannot integrate against it, and you cannot divide by it cleanly. It has to be simplified before step three. Twice a logarithm is the logarithm of the square. And the exponential of a logarithm is the thing itself.

12:37So the factor is x squared. A clean, ordinary function. That collapse is the teachable moment in this example, not the integration. Step three. y times x squared equals the integral of x times x squared, which is x to the fourth over four, plus a constant. Divide, and y is a quarter of x squared, plus the constant over x squared. Third one, and it comes with a data point. The derivative of y, plus y times the cotangent of x, equals twice x plus x squared times the cotangent.

13:19P is the cotangent. Its integral is the logarithm of the sine, so the factor is the exponential of that logarithm, which is the sine itself. Step three. y times the sine equals the integral of twice x plus x squared cotangent, all times the sine. Multiply through and that right hand side splits into two integrals. Twice x times the sine, and x squared times the cosine. Now do the second one by parts. It gives x squared times the sine, minus the integral of twice x times the sine.

14:01And that leftover is exactly the first integral, with a minus sign. The two annihilate. What is left is x squared times the sine, plus a constant. Divide by the sine. y equals x squared, plus the constant over the sine. Now the data point. y is nought at a quarter turn. The sine of a quarter turn is one, so nought equals a quarter of pi squared plus the constant, and the constant is minus a quarter of pi squared.

14:38The answer is x squared, minus a quarter of pi squared over the sine. And here is where care pays. There is a natural misreading in which the whole numerator goes over the sine: x squared minus a quarter of pi squared, all divided by the sine. That version passes through the data point too. It is nought at a quarter turn, exactly like the right one. And it solves the equation nowhere. At all five places tested its slope is wrong. Meeting the data point is not evidence that an answer is right.

15:19The last worked one hands you no equation at all. It hands you a sentence. The slope of the tangent equals the x coordinate added to the product of the two coordinates. Find the curve through the point where x is nought and y is one. Translate it phrase by phrase. The slope is the derivative of y. The x coordinate is x. The product of the two coordinates is x times y.

15:49So the derivative of y equals x plus x y. Move the y term across: the derivative of y minus x times y equals x. P is minus x. Its integral is minus half x squared, so the factor is the exponential of minus half x squared. Step three, integrate by substitution, divide, and apply the point. The answer is minus one, plus twice the exponential of half x squared.

16:24Now the argument I said the first worked example would settle. Step three gives you y times M equals an integral plus a constant. To read off y you divide by M, and division does not choose which terms it applies to. It applies to all of them. So y is the integral over M, plus the constant over M. There is a version of this formula that circulates with the constant left outside: y equals the integral over M, plus C on its own.

17:00That is a different function, and the difference is not cosmetic. Take the equation from the second example, where the right answer is a quarter of x squared plus the constant over x squared. The version with the constant outside was tested at five constants. It solves the equation for one of them and fails for the other four. The one it works for is nought, which is exactly why the slip survives a casual reading.

17:34One more trap, and it lives in the wording of questions rather than in the mathematics. Some of these problems state their slope condition through a size. The two coordinates added run ahead of the size of the slope by a fixed amount. A size has no sign. So that sentence is two equations wearing one coat, and it asks for one curve. Take the fixed amount to be five and the starting point to be where x is nought and y is five. Both branches are genuine linear equations, and both were solved and measured.

18:13The positive branch gives four minus x plus the exponential of x. The negative branch gives six minus x minus the exponential of minus x. Both pass through the same starting point. Both satisfy the sentence. And at every other place tested they are different curves. They agree at exactly one place, and it is the starting point itself, where the size of the slope is nought and a nought has no sign to read. That is why the point cannot decide between them.

18:52When a question is written that way, take a branch, say which one you took, and move on. So, what to carry away. Linear is a property you can decide, not a shape you recognise. The unknown and its derivative, each to the first power, each outside everything else. The method is one fold. Multiply by the exponential of the integral of P, and two terms become the one derivative they came from.

19:23Drop the constant when you find the factor. Never drop it at the end. And when you divide by the factor, the constant comes along.