Homogeneous equations, and the substitution that separates the variables for you

0:00The previous method wanted the right hand side to come apart into an x piece times a y piece. Here is one that will not. The derivative of y equals x plus two y, all over x minus y. Try to split it and you get nowhere. There is no factorisation to find, because there is not one there. And yet this equation is solvable, by a method that is barely longer than the last one.

0:30What it has instead of a factorisation is a symmetry, and the whole of this video is about seeing that symmetry and cashing it in. Here is the move that finds it. Take an expression in x and y, and replace both variables at once by scaled copies. Not x alone. Both, by the same factor. Watch what four different expressions do under that one operation. The first is y squared plus twice x y. Scale both, and every term picks up the factor squared, so the whole expression comes out as the factor squared times what it was.

1:13The second is twice x minus three y. Scale both, and the factor comes out once. The third is the cosine of y over x. Scale both, and the ratio is unchanged, so the expression is unchanged. The factor comes out to the power nothing. The fourth is the sine of x added to the cosine of y. Scale both, and nothing comes out at all. There is no power of the factor that will do it.

1:46That is the definition. An expression is homogeneous of degree n when scaling both variables multiplies the whole thing by the factor to the n. So the first has degree two, the second has degree one, the third has degree nought, and the fourth has no degree at all. A word of warning, because the same word is about to be used for two different things. You have already met a degree that meant the power on the highest derivative. This one means the power of the scaling factor.

2:21They share a word and they share nothing else. For the equation with a cosine of the derivative in it, the scaling degree is nought and the derivative degree is one. Two numbers, two ideas, one word. Now the definition that matters, and this is where attention pays. A first order equation qualifies for this method when its right hand side is homogeneous of degree nought. Only degree nought. A page has just been spent on degrees two, one and nought. Two of those three were scaffolding. They were there to show you what a degree is, not to tell you which equations qualify.

3:06A student who does not notice will spend the whole exercise testing for the wrong thing, and will pass equations the method cannot touch. And here is the reason, which is worth more than the rule. Degree nought says scaling both variables changes nothing. So two points that lie on the same ray from the origin, with the same ratio of y to x, give the same reading. Which means the expression does not really depend on x and y separately at all. It depends on one number: the ratio.

3:45Check that by measurement. Read a degree nought expression at eight places along a ray, and you get one value, eight times over. Do the same to the degree one expression and the readings scatter. The collapse is a property of degree nought, and of nothing else. If the only thing on the right is the ratio, then make the ratio the unknown. Write y as a new letter times x. The new letter is that ratio.

4:17Now differentiate, and here is the step that catches more students than any other in this topic. The new letter is a function of x. It is not a constant. So this is a product of two functions of x, and it needs the product rule. The derivative of y is the new letter, plus x times the derivative of the new letter. Measured at five places, that is exactly what the slope reads. And the wrong first line, treating the new letter as a constant so that the slope is just the letter itself, closes at none of them.

4:57Substitute both of those into the equation, and watch the x disappear. On the left, the new letter plus x times its derivative. On the right, the original expression evaluated at y equal to the letter times x. Because the expression has degree nought, that right hand side is a function of the new letter alone. Every x in it cancels. So what is left is x times the derivative of the new letter, equal to a function of the new letter, and that separates by the previous method.

5:34That cancellation is the whole method. And it happens precisely because the degree was nought. At degree one it does not happen. Hold the ratio fixed, vary x, and the reading changes. The x survives, and nothing separates. Which is the answer to why the condition is degree nought and not degree anything. So the method, once and for all, in three moves. Substitute, and use the product rule. Separate the result and integrate it, exactly as in the previous topic.

6:12And then put the ratio back. The answer has to be in the original variables, because that is what was asked. That last move is not optional and it is not decoration. An answer left in the new letter is an answer to a different question. Start with the shortest complete example, because the shape of the method shows better when the algebra is out of the way. x times the cosine of the ratio, times the derivative, equals y times the cosine of the ratio, plus x.

6:50Substitute. The left becomes x cosine of the letter, times the letter plus x times its derivative. The right becomes the letter times x times the cosine, plus x. Divide through by x, and the terms carrying the letter times the cosine appear on both sides and cancel. What is left is startlingly small. x, times the cosine of the letter, times the derivative of the letter, equals one. Separate: cosine of the letter, d letter, equals d x over x.

7:28Integrate: the sine of the letter equals the logarithm of x plus a constant. Put the ratio back: the sine of y over x equals the logarithm of x plus a constant. Three lines, and done. And that answer is not taken on trust here. Solve it for y, read it back as a curve, and measure its slope from difference quotients at five places. At every one of them the measured slope agrees with what the equation demands, to within a tolerance far tighter than the reading.

8:06Move the target by a fiftieth and it fails at all five. So the agreement is a measurement, not a courtesy. Now the one that is usually met first, which is a much longer piece of work. x minus y, times the derivative, equals x plus two y. The substitution goes in the same way, and after the x cancels the separated equation is one minus the letter, over one plus the letter plus the letter squared.

8:39That quadratic on the bottom is irreducible. Its discriminant is minus three, and that minus three is where every square root of three in the final answer comes from. The integral on the left needs a standard trick, and it is worth naming rather than performing silently. The numerator is split into two parts: a multiple of the derivative of the denominator, plus a leftover constant. The first part integrates to a logarithm, because that is what a derivative over its own function always gives.

9:19The second part, after completing the square on the bottom, is the standard form that integrates to an inverse tangent. Completing the square leaves three quarters, and the root of three quarters is what puts a square root of three inside the inverse tangent, in the denominator. And the constant in front works out to twice the square root of three. Both radicals, from one negative discriminant. There is one step in that working that tends to go past without comment, and it is worth a sentence.

9:57It is the step where two logarithms are combined into one. So here is the answer, in one sentence. A coefficient in front of a logarithm is a power inside it. Twice the log of something is the log of that something squared. That is all it is. Checked at six different pairs of numbers, it holds exactly. Put the ratio back and the answer is a logarithm of a symmetric quadratic in x and y, equal to twice the root of three times an inverse tangent, plus a constant.

10:37Checking an implicit answer like that takes a little care, because it is a relation, not a function. The derivation itself hands over the tidiest route. Solving the integrated equation for x gives x as an explicit rule in the new letter, and y is the letter times x. So the curve can be walked directly, with the letter as the parameter, and the slope assembled from two measured readings rather than asserted from either.

11:09Done that way, the answer solves the equation at every place read, and the relation holds one single value all along that curve. Which is what an implicit answer being right actually means. A different kind of question, and the most interesting picture in the topic. No equation is handed over. Instead, a family of curves is described by the slopes of its tangents: at every point the slope is the sum of the squares over twice the product.

11:45That is homogeneous of degree nought, so the method applies. The substitution gives twice the letter over one minus the letter squared, against the reciprocal of x. Integrating gives a logarithm of a difference of squares. And the family is x squared minus y squared equals a constant times x. Measured, those curves really do have the slopes they were described by, at every place read. Move the constant and they do not. The picture and the algebra agree.

12:22When a data point is given, there is an ordering that matters. The data point is given in x and y. The relation you have just integrated is in the new letter. So the ratio has to be undone before the point can be used. Apply it early, to the relation in the new letter, and you are reading the point wrongly. And here is something worth knowing. The exercise item that this warning is usually attached to has its data point at x equal to one.

12:56At x equal to one the ratio equals y, so the right reading and the wrong reading give the same constant. That item cannot catch the mistake it is warning about. Move to any point off that line and the two readings differ every time. Which is why the ordering is worth stating explicitly rather than learning from an example that cannot demonstrate it. One multiple choice question in the exercise exists purely to punish guessing, and it is a good test of whether the definition landed.

13:36Four equations are offered, and only one is homogeneous. Three of them look plausible until they are scaled. The first carries bare constants added on, and constants do not scale, so it has no degree at all. The third mixes degrees term by term, and has none either. The second is the interesting one. It is homogeneous, of degree minus one. A genuine fixed power. And it still does not qualify, because the condition is degree nought and not degree anything.

14:12The fourth is the answer: every term the same total degree, and the whole thing of degree nought. One last item, because it teaches the limit of the substitution. One question in that exercise is not a y equals a letter times x problem at all. It is written the other way up, as the derivative of x with respect to y. For those, the mirror substitution is the one that works: x equals a new letter times y.

14:47Run that on the item and it separates cleanly, and the answer is x plus y times an exponential of the ratio equals a constant. Measured at five places, it closes. The rule for choosing is the same test as always. Scale it, check the degree, and the form of the equation tells you which of the two substitutions to reach for. So, all of it together. Scaling both variables at once sorts expressions by the power they pick up, and that power is the degree.

15:25An equation qualifies for this method only when its right hand side has degree nought. The other degrees were illustrations. Degree nought means the expression depends only on the ratio, which is why making the ratio the unknown is the obvious move rather than a trick. The substitution needs the product rule, because the new letter is a function and not a constant. The x cancels precisely because the degree was nought, and that cancellation is the entire method.

16:01Put the ratio back at the end, and apply any data point after that, not before. And when the equation is written the other way up, use the mirror substitution. One test, checked by scaling, tells you which.