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

What to assume they know

• Order and degree, and what a first order equation is, from m01-t02
• General and particular solutions, and what a data point does to a family, from m02-t01
• Indefinite integration and the standard list, from Chapter 7 of this volume
• Integration by substitution, and by partial fractions, from the same chapter
• The logarithm and the exponential as inverses, and the laws of logarithms
• Reading a slope condition off a sentence and writing it as a derivative

What they should be able to do

• Recognise, from the shape of the right-hand side, whether an equation can be separated
• Carry out the separation as two divisions and state the condition each needs
• Integrate both sides and collect a single arbitrary constant on one side
• Absorb a doubled or logged constant into a fresh letter without losing generality
• Apply a given data point to turn a general answer into a particular one
• Translate a sentence about a slope, or about a rate, into an equation before solving anything
• Set up and solve a continuous-growth problem where the rate is proportional to the quantity present
• Say which constant functions a separation throws away, and check whether any of them solve the original equation
• Choose the right partial-fraction split when the x-side does not integrate on sight

Where it usually goes wrong

• "Separable means I can always split it." The right-hand side has to factorise into an x-piece times a y-piece. Nothing in Exercise 9.4 does, which is exactly why the next method exists, and a student who tries to separate the first item of that exercise will spend ten minutes proving it cannot be done.
• "Two integrals, so two constants." One constant is enough, because a constant on the left can be moved to the right and combined. The chapter takes this silently in every worked example; say it once, out loud, in section 4.
• "The constant I write down is the constant in the answer." Example 4 renames a doubled constant, Example 9 renames an exponentiated one, and both are legitimate because the letter stands for any value at all. A student who refuses to rename ends up with an answer that is correct and unrecognisable.
• "I divided by y, so I am finished." Dividing throws away the values of y that make the divisor zero, and one of them is often a solution. Three places in this topic hide one: Example 6 excludes y equal to zero, which solves the equation; Exercise 9.3 item three excludes y equal to one, which solves the equation; item two restricts y strictly between minus two and two, and the endpoints solve the equation. Each of the three checked by substitution. Example 4's exclusion is different — there the excluded value makes the right-hand side undefined, so nothing is lost. The chapter prints every one of these restrictions and remarks on none of them.
• "A modulus inside a logarithm is decoration." The chapter writes the modulus in Example 7 and in Exercise 9.3 items three, five, seven and ten. It is what lets the answer cover negative x as well as positive.
• "An applied problem is a different kind of question." Every one of items fifteen to twenty-two becomes an ordinary separable equation the moment the sentence is translated. The translation is the marked step; the solving is routine.
• "Rate of five per cent means multiply by five over a hundred once." Continuous growth means the rate at every instant is proportional to the amount present at that instant, which is a differential equation, not a multiplication. Example 9 turns on exactly this reading.
• "If the derivative sits inside a function, I cannot separate." Exercise 9.3 item thirteen has the derivative inside a cosine and separates perfectly once the cosine is inverted. What sits inside a function affects whether degree is defined, which is a different question from whether the equation can be solved.
• "The answer has to be y as a function of x." Most of the answers in this topic are relations, and the chapter is content to leave them so. Example 4's answer is never solved for y at all.

Questions to check understanding

• Decide whether a given equation separates, and say which factor is which
• Separate and integrate a given equation to a general solution — the form of Exercise 9.3 questions 1 to 10
• Apply a stated data point to reduce a general solution to a particular one — the form of questions 11 to 14
• Turn a sentence about the slope of a tangent into an equation, then solve it — the form of questions 17 and 18
• Set up and solve a growth problem where the rate is proportional to the amount present — the form of questions 20 to 22
• Name a constant function discarded by a given separation, and say whether it solves the original equation
• Choose a general solution from four options — the form of question 23
• Rewrite an arbitrary constant in a more convenient form and justify the change

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 one-line promise of §9.4 (Part II p. 306). A single sentence announces three methods and names the class of equation they serve. It is the only place in the chapter where the three are set beside one another, and it is worth the explanation's first fifteen seconds because it tells a student that everything from Part II p. 306 to p. 329 is one project with three branches.
• The set-up of §9.4.1 (Part II pp. 306–307), numbered one to four by the chapter. Equation one is the general first order first degree shape. Then the condition: if the right-hand side is a function of x times a function of y, the equation is of the separable kind. Then, provided the y-factor is not zero, the division; then the integration; then the closing statement that the two antiderivatives, set equal up to one constant, are the answer. Verified by working the four steps independently: the derivation is complete and nothing is asserted without the division being licensed first.
• Example 4 (Part II p. 307). A first derivative equal to a linear expression in x over a linear expression in y, with the value that would make the denominator vanish excluded in brackets. Separates in one move. Verified: integrating gives twice y minus half y squared on the left and half x squared plus x on the right, and doubling clears the halves to a circle-shaped relation with the doubled constant renamed. The renaming is the teaching point of this example, not the integration: the chapter explicitly writes the new letter as twice the old one, which is the first time in the chapter a student sees that an arbitrary constant may be reshaped freely.
• Example 5 (Part II p. 308). One plus y squared under the derivative on one side, one plus x squared on the other. Verified: both sides integrate to inverse tangents, and the answer is one inverse tangent equal to the other plus a constant. Use this as the shortest complete instance in the chapter — three lines from statement to general solution — and note the chapter's care in saying first that the y-factor cannot vanish.
• Example 6 (Part II p. 308). A first derivative equal to minus four x times y squared, with y equal to one when x is zero. Verified: separating gives the reciprocal of y squared against x, integrating gives minus the reciprocal of y equal to minus twice x squared plus a constant, and the data point fixes that constant at minus one, so the particular answer is the reciprocal of twice x squared plus one. Note for section 11: the chapter opens the solution with the words "if y is not zero", and the constant function y equal to zero does satisfy the original equation. The chapter never says so.
• Example 7 (Part II pp. 308–309). x times dy equals a quadratic in x times dx, and the curve is required through the point where both coordinates are one. Verified: dividing by x gives twice x plus the reciprocal of x, so y is x squared plus the log of the modulus of x plus a constant, and the given point forces that constant to zero. This is the cleanest place to say out loud that the general answer is a whole family and the point selects a member — the chapter says exactly that on Part II p. 309, in the only two sentences of the topic that draw the picture in words.
• The starred footnote (Part II p. 309, foot). The chapter defends treating dy and dx as separable symbols, attributing the flexibility of the notation to Leibniz and citing a named calculus text and page for the argument. It is the chapter's only footnote, and it arrives after the manoeuvre it licenses has already been used three times — Examples 4, 5 and 6 all separate before the reader is told why separating is allowed. Move it to section 3 in the explanation, where it belongs. Read the citation off the page before naming the authors.
• Example 8 (Part II pp. 309–310). No equation is given; a sentence says the slope at a point equals twice x over y squared, and asks for the curve through a stated point. Verified: the slope is the derivative, so y squared dy equals twice x dx, integration gives a third of y cubed equal to x squared plus a constant, the point fixes the constant at five, and the answer can be written as a cube root. This is the first item in the chapter where the translation from words is the whole difficulty, and Exercise 9.3 has eight more like it.
• Example 9 (Part II p. 310). Continuous growth of a sum of money at five per cent a year; how long to double. Verified: the rate proportional to the amount separates to a logarithm, the exponential constant is fixed by the opening balance, and the doubling time comes out as twenty times the natural log of two — which the chapter leaves in that exact form rather than evaluating. Keep it unevaluated; the teacher says "twenty times the natural logarithm of two".
• Exercise 9.3, questions 1 to 10 (Part II pp. 310–311). Ten general solutions. Verified, working added here on each: item one needs the half-angle identity before it separates and lands on a tangent of half x minus x; item two carries a square root over four minus y squared and gives an inverse sine, so the answer is a sine of x plus a constant, doubled; item three separates to a logarithm and gives one minus an exponential in minus x; item four pairs a squared secant over a tangent on each side and multiplies to a product of tangents; item five is a logarithm of a sum of two exponentials; item six is an inverse tangent against a cubic; item seven needs the substitution that turns the reciprocal of y log y into the log of a log; item eight is a pair of fourth powers in the denominator and gives a sum of two inverse fourth powers; item nine is an integration by parts; item ten pairs an exponential over one minus an exponential against a squared secant over a tangent. Item two's square root does not survive text extraction — see Notes.
• Exercise 9.3, questions 11 to 14 (Part II p. 311). Four particular solutions. Verified: item eleven needs a partial-fraction split of a cubic that factorises into a linear and a quadratic factor, and is the hardest piece of integration in the topic; item twelve needs a three-way split and yields logs; item thirteen has the derivative inside a cosine, so it is solved by taking an inverse cosine of the constant and integrating; item fourteen separates to a logarithm of a secant. Item thirteen deserves a mention: its degree is not defined, by the test of m01-t02, yet it sits in an exercise attached to a section whose heading announces first degree equations. It is solvable and it is legitimately here; it is not first degree.
• Exercise 9.3, questions 15 to 22 (Part II pp. 311–312). Eight problems posed in words: two slope conditions and a tangent condition; one solution curve through a point; an inflating balloon whose volume grows at a constant rate; two more compound-interest problems; and a bacterial count growing in proportion to itself. Verified on the balloon: volume is four thirds pi r cubed, its rate is constant, and the two given radii at the two given times fix both constants, so the radius after t seconds is a cube root of a linear expression in t. Verified on the two interest items: the first inverts the doubling relation to give a rate near seven per cent, and the second is a direct evaluation; both print the numerical constant the student needs inside the question, which tells you the chapter expects no calculator.
• Exercise 9.3 question 23 (Part II p. 312), multiple choice. A first derivative equal to an exponential of the sum of x and y. Verified: the exponential of a sum splits into a product, the equation separates, and the answer is the exponential of x plus the exponential of minus y equal to a constant, which is the first option.
• The Summary bullet for this method (Part II p. 336). One bullet, and it is descriptive rather than procedural: it says the terms in y stay with dy and the terms in x stay with dx. It does not name a condition, mention the arbitrary constant, or say what to do when the split is not obvious. Treat it as a mnemonic, not a method.

Figures to have open

• A two-bracket factorisation card for section 2, showing a right-hand side splitting into an x-only factor and a y-only factor, with an unfactorisable example struck through beside it. The shapes are the chapter's; the side-by-side treatment is added here.
• A family-and-member frame for section 7: several curves of Example 7's family drawn on one set of axes, with the member through the given point thickened and the point marked. The chapter draws nothing of the kind — it prints no figure at all — so this is an added construction. Reuse the frame built for m02-t01 section 10 so the two topics visibly share a spine.
• A lost-solution panel for section 11: the division step with the discarded constant function shown falling out of frame, then returning and being substituted into the original equation. Not in the book; the chapter prints the exclusions and never draws the consequence.
• A four-band table for section 12, built with the repo's DataTable component. Content is the shape of Exercise 9.3 as printed on Part II pp. 310–312; the banding is added here.
• 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 — the sentence naming the three methods, p. 306
• §9.4.1, the separable condition and the four numbered steps of the derivation, Part II pp. 306–307
• Examples 4 and 5, Part II pp. 307–308
• Examples 6, 7 and 8 with their data points, Part II pp. 308–310
• The starred footnote on the flexibility of the derivative notation, with its cited source, Part II p. 309
• Example 9, continuous growth of a principal, Part II p. 310
• Exercise 9.3, questions 1 to 23, Part II pp. 310–312
• Summary, the variable separable bullet, Part II p. 336