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

What to assume they know

• Separating variables and integrating, from m03-t01
• The product rule, for differentiating a product of two functions
• Partial fractions, and completing the square under an integral, from Chapter 7
• The standard integrals giving a logarithm and an inverse tangent
• Substitution inside an integral, and changing the variable back at the end
• Laws of logarithms, including turning a sum of logs into a log of a product

What they should be able to do

• Test a two-variable expression for homogeneity by scaling both variables
• State the degree of a homogeneous expression, and say what degree zero means
• Recognise that a first order equation qualifies for this method only when its right-hand side is homogeneous of degree zero
• Rewrite a degree-zero expression as a function of the single ratio of the two variables
• Perform the substitution, differentiate it with the product rule, and see the ratio-only equation that results
• Separate and integrate the resulting equation, then put the original ratio back
• Apply a data point after the substitution has been undone, not before
• Recognise the three shapes that turn up repeatedly: a ratio of two quadratics, a trigonometric function of the ratio, and an exponential of the ratio
• Say why the method works — the substitution is designed to make the ratio the only thing left

Where it usually goes wrong

• "Homogeneous means the degrees add up to zero." It means scaling both variables together multiplies the whole expression by a fixed power of the scaling factor — and for a differential equation the fixed power must be zero. Degrees two and one appear on Part II p. 312 only as illustrations, and neither qualifies an equation for this method.
• "This degree is the degree from the order-and-degree topic." It is not. There, degree was the power on the highest derivative; here it is the power of the scaling factor. The two words are the same and the two ideas share nothing. The Key terms table flags the collision, and it is worth ten seconds.
• "I substitute, solve, and I am done." The answer must go back into the original variables. The chapter says so explicitly at the end of the derivation, and every worked example ends with that step written out.
• "The new letter is a constant, so its derivative is zero." It is a function of x, which is why the substitution needs the product rule. This is the single most common wrong first line, and it kills the method outright.
• "Apply the data point as soon as I have integrated." The data point is given in x and y, so the ratio has to be undone first. Every particular solution in Exercise 9.4 is written this way.
• "An equation with x's and y's mixed together is homogeneous." Test it. Exercise 9.4 item seventeen exists precisely to punish guessing, and three of its four options look plausible until scaled.
• "If the substitution does not tidy it, I have made an arithmetic slip." Sometimes the equation is genuinely not homogeneous — and sometimes, as in item ten of Exercise 9.4, the right substitution is the mirror one. Check degree zero first; that answer tells you which of the two to use.
• "The chapter spells it one way, so I should too." The chapter spells it two ways, and one of them appears inside the defining sentence. See section 11 and the Notes.

Questions to check understanding

• Test a given expression for homogeneity and state its degree
• Decide whether a given equation qualifies for this method, and justify the answer by scaling
• Solve a homogeneous equation by substitution — the form of Exercise 9.4 questions 1 to 10
• Find a particular solution given a data point — the form of questions 11 to 15
• Choose the homogeneous equation from four options — the form of question 17
• Rewrite a degree-zero expression as a function of a single ratio
• Explain in two sentences why the substitution leaves an equation in the new letter alone
• Given a worked solution with the final replacement step missing, supply it

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The four test expressions (§9.4.2, Part II p. 312). A quadratic in y with a cross term; a linear expression in both; a cosine of the ratio of y to x; and a sine of x added to a cosine of y. Each is scaled and the results are displayed. Verified by scaling each independently: the first picks up the square of the factor, the second the factor itself, the third nothing at all, and the fourth cannot be written as any power of the factor times the original. Build all four and scale them together — this is the single clearest thing in the section and it is entirely visual.
• The definition of a homogeneous expression (Part II p. 312). Scaling both variables by a nonzero factor multiplies the expression by a fixed power of that factor. The chapter then reports the degrees of the first three as two, one and zero, and says the fourth has none. Verified against the four scaled lines above.
• The rewriting as a function of the ratio (Part II p. 313). Every one of the first three is shown twice, once with the x-power pulled out and once with the y-power pulled out, and the failure of the fourth is shown both ways too. This is the step that makes the substitution obvious rather than magical, and an explanation that skips it leaves the substitution looking like a trick.
• The definition of a homogeneous equation (Part II p. 313). A first derivative equal to an expression in x and y qualifies when that expression is homogeneous of degree zero. Only degree zero. The section has just spent a page on degrees two, one and zero, and a student who has been reading attentively will need telling that the other degrees were scaffolding.
• The derivation of the method (Part II pp. 313–314), numbered one to six by the chapter. Substitute y equal to a new letter times x; differentiate with the product rule to get the new letter plus x times its derivative; substitute into the equation; the x-terms cancel to leave x times the derivative of the new letter equal to a function of the new letter alone; separate; integrate; and put the ratio back. Verified step by step: the cancellation is exact and needs nothing beyond degree zero, which is the whole reason the condition is degree zero and not degree anything.
• Example 10 (Part II pp. 314–316). A linear expression times the derivative equal to another linear expression. Verified end to end, and it is the longest single derivation in the chapter: the substitution reduces it to a ratio of a linear expression in the new letter over an irreducible quadratic in it; splitting the numerator into the derivative of the quadratic plus a constant gives a logarithm plus an inverse tangent after completing the square; and the ratio put back gives a logarithm of a symmetric quadratic in x and y equal to a multiple of an inverse tangent plus a constant. The multiple carries a square root of three, and so does the denominator inside the inverse tangent. Both radicals read on Part II p. 316. Working added here reproduces the printed answer exactly.
• The chapter's own "Why?" (Part II p. 316). The working carries a printed parenthetical question at the step where two logarithms are combined. It is the only place in the chapter where the reader is asked to supply a justification: the two logs combine because a coefficient in front of a log becomes a power inside it.
• Example 11 (Part II pp. 316–318). A cosine of the ratio multiplying the derivative. Verified: the substitution collapses the whole right-hand side to the reciprocal of a cosine, so the separated equation is a cosine of the new letter against the reciprocal of x, integration gives a sine equal to a logarithm, and the answer is a sine of the ratio equal to the log of a constant times x. This is the example to teach with, not Example 10 — the algebra is three lines and the shape of the method is completely exposed.
• Example 13 (Part II pp. 319–320). No equation is handed over; a family of curves is described by the slope of its tangents, and the answer is asserted in advance. Verified: the slope is a ratio of a sum of squares to twice the product, the substitution gives twice the new letter over one minus its square against the reciprocal of x, integration gives a logarithm of a difference of squares, and the result rearranges to the printed relation. Note that the chapter states the answer in the question, so this is a verification dressed as a derivation — useful, because a student can check progress at every line.
• Exercise 9.4, questions 1 to 10 (Part II p. 321). Ten to show homogeneous and solve. Verified, working added here on each: item two is by far the simplest and should be worked first — the substitution reduces it to a bare reciprocal of x, so the answer is x times a logarithm plus a constant. Item one separates to a ratio of two linear expressions in the new letter; item three gives an inverse tangent against a logarithm; item four gives a circle through the origin with its centre on the x-axis; item five needs a partial-fraction split with a square root of two in it; item six carries a square root over the sum of two squares and lands on the standard integral giving an inverse hyperbolic shape, written by the chapter's own machinery as a logarithm; item seven is the longest statement in the chapter; item eight gives the tangent of half the ratio times x equal to a constant; item nine involves a logarithm of the ratio; item ten is not solved by this substitution at all — it is a dx-over-dy item and belongs to m03-t04.
• Exercise 9.4, questions 11 to 15 (Part II p. 321). Five particular solutions, each with a data point. Verified: item eleven pairs an inverse tangent with half a logarithm; item twelve rearranges to a relation between x squared y and a linear combination, with the constant a third; item thirteen has a squared sine of the ratio and its data point is a quarter of pi; item fourteen has a cosecant of the ratio and gives a cosine of the ratio equal to a logarithm plus one; item fifteen separates to the reciprocal of the square of the new letter and gives twice x over y equal to one minus a logarithm. Items ten and thirteen are lost entirely by text extraction — see Notes.
• Exercise 9.4 questions 16 and 17 (Part II pp. 321–322), both multiple choice. Verified: item sixteen asks which substitution suits the dx-over-dy form and the answer is x equal to the new letter times y, the third option — that item belongs to m03-t04. Item seventeen asks which of four equations is homogeneous and the answer is the fourth, the only one whose every term has the same total degree; the first carries bare constants, and the second and third mix degrees.
• The Summary bullet for this method (Part II p. 336). One bullet, and unusually it is complete: it gives both the dy-over-dx and the dx-over-dy forms and states the degree-zero condition for each. It does not mention the substitution. A student revising only from the Summary can recognise one of these equations and cannot solve it.

Figures to have open

• A four-expression scaling build for section 2. Each expression appears, both variables are replaced by scaled copies, and the factor is shown coming out in front — twice over for the first, once for the second, not at all for the third, and refusing to come out for the fourth. This is the load-bearing image of the topic. The expressions are the chapter's; the figure is added here.
• A cancellation frame for section 6: the substituted equation with the x's ringed and then removed, so that a student sees the reason for the degree-zero condition rather than being told it.
• A slope-field-free family drawing for section 10: several members of Example 13's family on one set of axes with tangent segments marked at a few points. The chapter draws nothing of the kind — it prints no figure at all — so this is an added construction and it is the most valuable picture in the topic.
• A spelling comparison card for section 11 setting the printed heading against the printed defining sentence.
• A three-band table for section 12, built with the repo's DataTable component. Content is the shape of Exercise 9.4 as printed on Part II pp. 321–322; the banding is added here.
• No textbook figure can be redrawn: this chapter prints none. All thirty-eight pages were opened as page images.

Where this sits in the book

• NCERT Class 12 Mathematics, Part II, Chapter 9 "Differential Equations", §9.4.2 — the four test expressions, the scaling and the definition, pp. 312–313
• The rewriting of each expression as a power times a function of the ratio, Part II p. 313
• The definition of a homogeneous equation and the six-step derivation, Part II pp. 313–314
• Example 10 with its completed square and its two radicals, Part II pp. 314–316
• Example 11, Part II pp. 316–318
• Example 13, the family described by its slopes, Part II pp. 319–320
• Exercise 9.4, questions 1 to 17, Part II pp. 321–322
• Summary, the homogeneous bullet, Part II p. 336