Pushing all the y's one side and all the x's the other, then integrating

0:00There are three standard methods for first order equations, and they are all aimed at the same target. One class of problem, three ways in. This video is the first of them, and it is the easiest by a wide margin. It is three moves of bookkeeping. See that the right hand side comes apart. Divide. Integrate. That is genuinely all of it, and you could be solving problems within a minute.

0:30Which is exactly why it is worth slowing down, because underneath those three moves are two things that almost never get said out loud. One is the licence: why you are allowed to do the middle move at all. The other is the cost: what the middle move quietly throws away. Start with the shape. An equation of this kind says the derivative of y equals some expression built from x and y together.

1:01It is called separable when that expression comes apart into two pieces multiplied: a piece that only knows about x, times a piece that only knows about y. Not added. Multiplied. That distinction is the whole of the condition. And notice what separable is not. It is not a property of how the equation looks on the page. It is a property of the function on the right. So how would you know? You could hunt for the factorisation, and if you find one, you are done.

1:38But there is a test that needs no hunting at all. Read the right hand side at four places, arranged as a little rectangle. If the expression really is an x piece times a y piece, then the product of the two opposite corners equals the product of the other two. The x pieces cancel against each other, and so do the y pieces. Run that across a grid. Seven of the right hand sides in this topic pass it at every corner tried.

2:11And two shapes fail it. That failure is not bad luck, and it is not a student being careless. Those two are the reason the second and third methods exist. No amount of rearranging will separate them, and knowing that early saves you ten minutes of trying. Now the move itself, and the thing that deserves saying before you use it rather than after. You are about to take the derivative of y with respect to x, and treat the top and the bottom as two separate quantities you may move to opposite sides.

2:50Written down, that looks like an abuse. The derivative is a limit. It is one indivisible object, not a fraction. The licence comes from the notation being built to survive exactly this. The symbols were chosen so that the bookkeeping of moving them about tracks a real substitution underneath. What is really happening is a change of variable inside an integral, and the notation is arranged so that the shorthand and the substitution agree.

3:24So the move is legitimate. But you should meet the reason before you meet the manoeuvre, because a step you have used three times before anyone justifies it is a step you will use in places where it does not hold. Here is the separation, done slowly. The right hand side is an x piece times a y piece. Divide both sides by the y piece. Now every y sits on the left with its derivative, and every x sits on the right.

3:58Multiply through by the small change in x. The left side is now something in y against the change in y, and the right is something in x against the change in x. Two moves, and each one needed something to be true. The first needed the y piece not to be zero, because you cannot divide by nothing. Hold on to that condition. It looks like a technicality, and it is the single most consequential sentence in this video.

4:33Now integrate both sides. Two integrals, so surely two arbitrary constants? No. One is enough, and here is why. Put a constant on each side if you like. Then move the left one across. What arrives on the right is the difference of the two, and a difference of two arbitrary constants is just another arbitrary constant. Checked across four different pairs, the two constant version names exactly the same curve as the one constant version carrying their difference. Every time.

5:11And that is a real comparison rather than a restatement, because two pairs with different differences do name different curves. So the second letter buys you nothing. One constant, on one side, and you are done. Take a concrete one. The derivative of y equals a linear expression in x, over a linear expression in y. It separates in a single move: bring the y expression to the left, the x expression to the right.

5:45Integrate. The left gives twice y less half of y squared. The right gives half of x squared plus x, and one constant. Now there are halves everywhere, which is untidy. Multiply the whole line by two. And here is the moment worth watching. Multiplying by two also doubles the constant. So the constant in the tidy version is twice the constant in the untidy one. The usual move is to shrug and call the doubled thing a new letter. That shrug is correct here. But it is a shrug worth examining, because the same shrug is not always correct.

6:29A constant is renamed whenever it changes shape. Doubling it. Exponentiating it. Taking it inside a logarithm. The question you should ask each time is simple: can the new letter reach everything the old one could? For doubling, yes. Name any value you want the new letter to take, positive, negative, or nought, and halving it gives you the old constant that produces it. Nothing is lost. For exponentiating, no. An exponential is always strictly positive.

7:06Take six target values, three positive and three not. All six are reachable by doubling. Only the three positive ones are reachable by exponentiating. So when a constant gets exponentiated, the new letter silently loses the whole of the negative half line, and it loses nought. Remember that nought. It comes back later in this video, carrying a solution with it. A second example, and the shortest complete one there is.

7:40One plus y squared sits under the derivative on one side, and one plus x squared on the other. Both sides integrate to inverse tangents. The answer is one inverse tangent equal to the other, plus a constant. Three lines from start to finish. Read that answer back as an actual curve and measure its slope, and it matches the right hand side at every place checked, for every constant tried.

8:11And the inverse tangent and the tangent used to check it were both computed from series, not looked up in a table. The check is a measurement. So far every answer has been a family. A whole set of curves, one for each value of the constant. One piece of data collapses it to one curve. Here is a family: y equals x squared, plus the logarithm of the size of x, plus a constant.

8:43Ask for the member passing through the place where both coordinates are one. The squared term gives one, the logarithm of one is nought, so the constant must be nought. That member passes through the point. The neighbouring members, at constant one, at minus one, at a half, all miss it. That is the whole relationship between a general answer and a particular one. The general answer is the family. The data point is what selects a member.

9:18Sometimes no equation is handed to you at all. You get a sentence. The slope of the curve at any point is twice x over y squared, and the curve passes through a stated place. The translation is the only hard part. The slope is the derivative. So y squared against the change in y equals twice x against the change in x. Integrate: a third of y cubed equals x squared plus a constant. The stated point fixes that constant at five, and the answer can be written as a cube root.

9:57Checked by measurement, that curve closes the equation at every place read, and the neighbouring constants miss the point it was built to pass through. Once the sentence is translated, the solving is routine. It is the translating that is the marked step, and there is no way to practise it except on sentences. Now the part this video exists for. Go back to the very first division. To separate, you divided by the y piece. And to divide by something, it must not be nought.

10:35So every separation quietly sets aside the values of y that make that piece zero. They are excluded, in brackets, and everyone moves on. Here is the question almost nobody asks. Those excluded values are constant functions. Do any of them solve the original equation? A constant function has a slope of nought everywhere. So it solves the equation exactly when the right hand side is also nought everywhere along it.

11:08Take the equation whose right hand side is minus four x times y squared. The separation sets y away from nought. Put the constant function nought back in. Its slope is nought. The right hand side, with y at nought, is also nought. At every one of the seventeen places checked. It solves it. And it is nowhere in the general answer, because that answer is a reciprocal, and a reciprocal is never nought.

11:41A whole solution, thrown away by a division, and never mentioned again. That was one. Here is a second, and this one shows you the exact machinery of the loss. Take the equation where the derivative of y equals one less y. Separating divides by one less y, so y equal to one is set aside. Check it: the constant function one has slope nought, and one less one is nought. It solves the equation.

12:13Now solve properly. Integrating gives a logarithm, and rearranging gives one less y equal to a constant times an exponential in minus x. And where did that constant come from? It came from exponentiating the constant of integration. Which is where the solution died. If that constant could be nought, the family would give y equal to one back to you. But it came from an exponential, and an exponential is never nought.

12:47This is the point from earlier arriving. The renaming is not cosmetic. Exponentiating a constant loses nought, and nought was carrying a solution. A third case, to show it is not a coincidence. The right hand side is the square root of four less y squared. That is real only between minus two and two, and the separation holds y strictly inside. Both endpoints are set aside. Both of them solve it. At y equal to two, four less four is nought, the root of nought is nought, and a constant function has slope nought. Same at minus two.

13:32So across three equations, four discarded values, and every one of the four is a genuine solution that the method threw away. And now the control, because a rule with no exceptions is usually a rule that has not been tested. Go back to the very first worked example, with the linear expression on the bottom. It excluded y equal to two as well. Put y equal to two back in. The bottom becomes nought, and the right hand side is not a number at all there.

14:08So that exclusion loses nothing. There was never a solution there to lose. Four of the five exclusions in this topic hide a solution. The fifth does not. Telling those apart takes one substitution, and it is one substitution nobody performs. Turn to where this method earns its keep. A sum of money grows continuously at five per cent. Not five per cent added once at the end of the year.

14:41Continuously means the rate of growth at every instant is proportional to the amount present at that instant. That sentence is a differential equation, not a multiplication. It separates immediately. The amount against its change on one side, the time on the other. Integrating gives a logarithm, and undoing it gives an exponential. How long to double? Set the amount to twice what it started as. The starting amount cancels on both sides, and what is left is twenty times the natural logarithm of two.

15:22Leave it in that form. It is exact, and evaluating it to a decimal loses that. And the cancelling is the real content. Checked against three different opening amounts, all three double in exactly the same time. Proportional growth does not care how much you started with. One more, because it looks harder than it is. A balloon is inflated so that its volume grows at a constant rate. What happens to the radius?

15:55The volume is a fixed multiple of the radius cubed. If the volume grows at a constant rate, then the cube of the radius grows at a constant rate. So the cube of the radius is linear in the time, and the radius itself is a cube root of something linear in the time. Measured directly, the rate of change of the cube comes back as the same number at every time it is read. Not four numbers. One.

16:26And that is derived without ever naming the constant sitting in front of the volume, which is the tidy part. Two readings of the radius at two times are enough to pin both constants, and the curve then passes through the pair it was built from. A loose end worth tying, because it looks like a contradiction. This whole family of methods is introduced for equations of first order and first degree.

16:57But degree only exists when the equation is a polynomial in its derivatives. Put the derivative inside a cosine, and there is no degree to speak of. And yet such an equation separates perfectly. Invert the cosine, and the derivative equals a constant, so y is a straight line. Read the equations rather than the heading: a plain first derivative has a degree, the same derivative inside a cosine has none, and both are of order one.

17:32Checked by measurement, the straight line closes the equation for every constant tried. The cosine of the measured slope equals the value on the right. So not first degree and not solvable are two different sentences, and only the first of them is true here. So, all of it together. Separable means the right hand side is an x piece times a y piece, multiplied, and there is a test that settles it without hunting for the factorisation.

18:07Moving the top and bottom of a derivative to opposite sides is legitimate, and the reason belongs before the move rather than after it. Two integrals need only one constant, because a difference of two arbitrary constants is another arbitrary constant. A renaming is not always free. Doubling a constant reaches everything. Exponentiating one loses the negative half line and loses nought. And the big one. Every separation divides by something, and dividing sets aside the values that make it zero. Four of the five values set aside in this topic are constant functions that solve the original equation. The fifth is not a number there at all.

19:00One substitution tells those apart. It costs about ten seconds, and without it the method hands you an answer that is not the whole answer. Which is the honest summary of the easiest technique here: three moves of bookkeeping, one licence worth understanding, and one habit of checking what the division threw away.