The mirror-image form, where x is treated as the dependent variable

What to assume they know

• The homogeneous method and its substitution, from m03-t02
• The integrating factor and the three-step recipe, from m03-t03
• Which symbol is the dependent variable and which the independent, from m01-t01
• The reciprocal relation between the two derivatives of a one-to-one relation
• Integration by parts, and substitution inside an integral
• The standard integral giving an inverse sine, for the closing choice item

What they should be able to do

• Recognise when an equation is easier to read with x as the unknown
• Swap the roles of the two letters and rewrite an equation accordingly
• Apply the homogeneous substitution in its mirrored form, with x as the multiple
• Apply the integrating-factor recipe in its mirrored form, integrating against the other letter throughout
• Decide, from the shape of a printed equation, which of the two orientations will be less work
• Resist the trap of choosing the orientation the equation is written in rather than the one it needs
• Complete a solution left in the mirrored orientation, without converting back
• Recall that the revision page names this form for one of the two methods and not for the other

Where it usually goes wrong

• "y is always the unknown." Which letter is the unknown is a choice, not a fact about the symbols. The chapter's own opening section names the two roles and says nothing about which letter has to play which, and this topic is where that becomes practical.
• "Inverting the derivative is illegal." Turning the ratio over is the whole move, and the chapter's footnote on Part II p. 309 is the licence for treating the two pieces as separable symbols. Where the derivative is nonzero the two ratios are reciprocals.
• "The equation is written with dy on top, so I solve for y." Look at which letter the equation is linear in, not at how it is written. Exercise 9.5 items ten to twelve are all written the forward way and all need mirroring; Miscellaneous Exercise item ten is written the mirrored way and needs the forward method. Both traps are printed, and they point in opposite directions.
• "I mirror it, solve it, and then convert the answer back." There is nothing to convert. A relation between x and y is a relation between x and y whichever letter was called the unknown. Every mirrored example in the chapter stops as soon as the relation is found.
• "The mirrored factor is the same factor." It is built by integrating the first coefficient against the other letter. Getting this wrong is the single most common error in the mirrored items, and it produces a factor that is a function of the wrong variable.
• "The substitution is the same substitution." It is the mirror: the new letter multiplies the other variable. Exercise 9.4 question 16 exists entirely to test whether a student knows which way round it goes, and its first option is the unmirrored one.
• "If the substitution works, the equation was homogeneous." Not always. Miscellaneous Exercise item eight yields to the mirrored substitution and is not homogeneous of degree zero — checked by scaling. The substitution is a tool that happens to be wider than the class the chapter defines it for, and a student who reasons backwards from the tool to the class will be wrong.
• "The revision page will remind me." For the homogeneous half it will. For the linear half it will not — see section 11 — and four exercise items depend on the half it drops.

Questions to check understanding

• Given an equation, say which letter it is linear in and which orientation to take
• Rewrite a given equation with the roles of the two letters exchanged
• Solve a mirrored homogeneous equation by the mirrored substitution — the form of Exercise 9.4 question 10
• Choose the correct substitution for the mirrored homogeneous shape — the form of Exercise 9.4 question 16
• Compute the integrating factor for a mirrored linear equation — the form of Exercise 9.5 question 19
• Solve a mirrored linear equation in full — the form of Exercise 9.5 questions 10 to 12
• Choose the correct general solution for the mirrored linear shape — the form of Miscellaneous Exercise question 14
• Given an equation printed with the mirrored derivative, decide whether it actually needs mirroring

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 pairing on Part II p. 313. Before either mirrored method appears, the chapter has already shown every one of its homogeneous test expressions factorised both ways — once with a power of x pulled out and a function of y over x left behind, and once with a power of y pulled out and a function of x over y left behind. Verified against all four expressions. This is the best possible opening for the topic and it costs nothing: the symmetry is printed, a page before it is used.
• The boxed Note (Part II p. 314). Four lines inside a tinted box. If the equation comes with the other derivative on the left and the right side is homogeneous of degree zero, then substitute x equal to a new letter times y and proceed as before. The note carries no section number, so a sweep looking for one will not see it, and it is the only place in the chapter where the mirrored homogeneous substitution is stated. Read closely.
• Example 12 (Part II pp. 318–319). Two terms, each carrying an exponential of the ratio of x to y, one against dx and one against dy, with a data point where x is zero and y is one. Verified end to end: rearranged as the mirrored derivative, the right side is homogeneous of degree zero; the substitution and the product rule give the new letter plus y times its derivative; everything collapses to twice an exponential of the new letter against the reciprocal of y with a minus sign; integrating gives twice that exponential plus a logarithm of the modulus of y equal to a constant; and the data point makes the constant two. This is the cleanest mirrored example in the chapter and should carry section 4 alone.
• Exercise 9.4 question 10 (Part II p. 321). A bracket holding one plus an exponential of the ratio, against dx, added to that same exponential times a bracket holding one minus the ratio, against dy. The text layer of that page loses this item completely — it leaves a bare item number and white space. Recovered at 300 dots per inch. Verified: it is a mirrored homogeneous item, it is the only one of the ten in that band, and an inventory built from extracted text would have missed it entirely.
• Exercise 9.4 question 16 (Part II p. 321), multiple choice. It states the mirrored homogeneous shape and asks which substitution solves it. Verified: the answer is x equal to a new letter times y, the third option; the first option is the unmirrored substitution and is the distractor the whole item is built around. Note that the stem repeats the printed slip from §9.4.3 — see Notes.
• "Another form" of the linear shape (Part II p. 322). Half a page: the mirrored standard shape, the condition that both its coefficients depend on the other letter alone, and two printed instances — one with a constant coefficient and a cosine on the right, one with a coefficient carrying a minus sign over the other letter and a product of a square and an exponential on the right. The minus sign read at 300 dots per inch. This paragraph has no heading of its own, numbered or otherwise; it simply follows the forward case.
• The mirrored recipe (Part II p. 324). One sentence and one displayed line, tucked under the same unnumbered heading as the forward recipe. The factor is an exponential of the integral of the first coefficient against the other letter, and the solution is the unknown times the factor equal to the integral of the second coefficient times the factor, plus a constant. The exponent of the factor is printed without its integral sign — see Notes.
• Example 16 (Part II p. 326). y times dx, minus a bracket holding x plus twice y squared, times dy. Verified: dividing by y and by dy gives the mirrored standard shape with first coefficient minus the reciprocal of y; the factor is an exponential of minus a logarithm, which simplifies to the reciprocal of y; the integral on the right collapses to a constant times y; and the answer is x equal to twice y squared plus a constant times y. Six lines from statement to answer — this is the item to use if the explanation is short of time.
• Example 22 (Part II p. 333), in the Miscellaneous Examples. A bracket holding an inverse tangent of y minus x, times dy, equal to a bracket holding one plus y squared, times dx. Verified: rearranged, it is the mirrored standard shape with first coefficient the reciprocal of one plus y squared and second coefficient an inverse tangent over the same thing; the factor is an exponential of the inverse tangent; the integral on the right yields to the substitution that sets the inverse tangent equal to a new letter, then to integration by parts; and the answer is x equal to the inverse tangent minus one, plus a constant times the exponential of minus the inverse tangent. This is the hardest item in the topic and the only one that combines the mirrored recipe with two integration techniques.
• Exercise 9.5 questions 10, 11 and 12 (Part II p. 329). Three of the twelve general-solution items are mirrored, and none of them is flagged as such. Verified, working added here: item ten inverts to a mirrored shape with constant first coefficient and gives a relation between x, y and an exponential of y; item eleven gives a factor equal to y and an answer with a cube of y over three; item twelve gives a factor equal to the reciprocal of y and an answer with three y squared. A student who does not know to invert cannot start any of the three, because each is linear in x and non-linear in y.
• Exercise 9.5 question 19 (Part II p. 329), multiple choice. Printed already in the mirrored orientation and asking only for the factor. Verified: dividing by the bracket gives a first coefficient of y over one minus y squared, whose integral against y is minus half a logarithm, so the factor is the reciprocal of the square root of one minus y squared — the fourth option. The four options differ only in a sign inside a bracket and in whether a radical is present, so all four were read at 500 dots per inch before this answer was fixed.
• Miscellaneous Exercise questions 8, 13 and 14 (Part II pp. 334–335). Verified: item eight is solved by the mirrored substitution — see the warning in Notes; item thirteen reduces to the simplest separable equation in the chapter and its answer is a straight line through the origin, the third option; item fourteen states the mirrored linear shape and asks for its general solution, and the answer is the third option, the only one that pairs the correct unknown with the correct letter of integration. All four options of item fourteen print their integral signs correctly, which is how the omission on Part II p. 324 is known to be a slip rather than a convention.
• Miscellaneous Exercise question 10 (Part II p. 335). Written with the mirrored derivative on the page and not a mirrored problem at all. Verified: multiplying out and inverting gives an ordinary forward linear equation whose first coefficient is the reciprocal of a square root of x, so the factor is an exponential of twice that square root, the integral on the right collapses to a constant, and the answer is the unknown times that exponential equal to twice the square root plus a constant. This is the exact reverse of the trap in section 10 and the two should be taught together.
• The Summary (Part II p. 336). Counted and read on the printed page. Eight bullets. The homogeneous bullet does carry the mirrored form — it prints both derivatives and applies the degree-zero condition to each. The linear bullet does not — it prints the forward shape only, names the condition on its two coefficients, and stops. So the chapter's revision page carries this topic forward for one of its two halves and drops the other — while Exercise 9.5 sets four items needing the dropped half, and the Miscellaneous Exercise adds a fifth.

Figures to have open

• A role-swap movement for section 1: one equation with the two letters labelled as unknown and as the variable it is differentiated against, and the two labels then exchanging. This is the load-bearing image of the topic, and it has to be built from nothing — the chapter draws no picture at all.
• A mirrored-shape comparison for section 5, with the forward standard shape above and the mirrored one below and each changed symbol highlighted. The two shapes are the chapter's; the alignment is added here.
• A two-column orientation table for section 9, built with the repo's DataTable component: three printed equations, which letter each is linear in, and which orientation is less work.
• A dropped-bullet comparison for section 11, setting the Summary's homogeneous bullet beside its linear bullet so the asymmetry is visible in one frame.
• A gathering table for section 12, also DataTable, in two bands. The first band holds what needs this topic: the boxed note, the second-form paragraph, the mirrored recipe line, Examples 12, 16 and 22, Exercise 9.4 items ten and sixteen, Exercise 9.5 items ten to twelve and nineteen, and Miscellaneous Exercise items eight and fourteen. The second band holds the two that only look as though they do: Miscellaneous Exercise item ten, which is printed mirrored and solves forward, and item thirteen, which is printed as a ratio of differentials and simply separates. Every row carries its Part II folio.
• No textbook figure can be redrawn: this chapter prints none. All thirty-eight pages were opened as page images and the only artwork in the chapter is the portrait and QR code on Part II p. 300.

Where this sits in the book

• NCERT Class 12 Mathematics, Part II, Chapter 9 "Differential Equations", §9.4.2 — the two-way factorisation of every test expression, p. 313
• The boxed Note giving the mirrored homogeneous substitution, Part II p. 314
• Example 12 with its data point, Part II pp. 318–319
• Exercise 9.4 questions 10 and 16, Part II p. 321
• §9.4.3 — the second form of the linear shape and its two instances, Part II p. 322
• The mirrored recipe line under the unnumbered heading, Part II p. 324
• Example 16, Part II p. 326; Example 22, Part II p. 333
• Exercise 9.5 questions 10 to 12 and 19, Part II p. 329
• Miscellaneous Exercise questions 8, 10, 13 and 14, Part II pp. 334–335
• Summary, the homogeneous and linear bullets compared, Part II p. 336