Multiplying through by an integrating factor to make the left side a single derivative

What to assume they know

• Separating variables and integrating, from m03-t01
• The product rule, forwards and read backwards as a pattern to recognise
• Integration by parts, including the case where the integral returns to itself
• The exponential of a logarithm, and the rule that turns a coefficient in front of a logarithm into a power inside it
• Standard integrals for the cotangent, the secant and the tangent
• Reading a slope condition off a sentence and writing it as a derivative

What they should be able to do

• Recognise the standard shape and name its two coefficient positions
• Rearrange an equation into that shape, dividing through when necessary
• Say why multiplying by a well-chosen function of x makes the left side collapse into one derivative
• Derive the condition that function must satisfy, and solve it
• Compute an integrating factor and simplify it, including the case where an exponential of a logarithm cancels
• Write the solution in the safe form: the unknown times the factor equals the integral of the other coefficient times the factor, plus a constant
• Apply a data point to fix that constant
• Translate a sentence about a slope into an equation of this shape and solve it
• Say why a constant of integration may be dropped when the factor is being found, but never when the solution is being written

Where it usually goes wrong

• "Linear means the graph is a straight line." It means the unknown and its derivative each appear to the first power and never inside another function. The chapter never says this, which is exactly why a student will reach for the graph. Nothing about the solutions of these equations is straight.
• "The coefficients can involve y." They cannot. The moment y appears in either coefficient position the equation leaves this class, and the integrating factor cannot be computed, because integrating the first coefficient with respect to x requires it to be free of y.
• "I can compute the factor before rearranging." The derivative must stand alone with coefficient one. Example 15 and Exercise 9.5 items five, six, eight and eighteen all need a division first, and reading the coefficients off the unrearranged equation gives the wrong factor every time.
• "I left out the constant when finding the factor, so my answer is wrong." Dropping it there is legitimate: a different constant multiplies the factor by a fixed nonzero number, and that number cancels from the final line. Dropping it at the end is fatal. The chapter drops it in the first place and keeps it in the second, and never explains the asymmetry.
• "The arbitrary constant just sits on the end of the answer." It sits on the end of the line before you divide by the factor, so dividing carries the constant along with everything else. Example 14 shows this correctly — its constant is multiplied by an exponential. The displayed formula on Part II p. 323 shows it incorrectly. See section 11.
• "An exponential of a logarithm has to be left as it is." It simplifies, and every worked example in this topic depends on the simplification. Example 15 turns an exponential of twice a logarithm into a square, and Example 17 turns an exponential of a logarithm of a sine into the sine.
• "If the integral on the right will not go, I have the wrong factor." Sometimes the integral is simply hard. Example 14's returns to itself and needs the by-parts trick; Example 17's needs two integrals to cancel. Neither is a sign of a wrong factor.
• "This method and the separable method compete." They do not. An equation of this shape with the second coefficient equal to zero also separates, and either route gives the same answer. The three methods overlap; the chapter never says so, and a student who thinks each equation has exactly one legal method will get stuck deciding.

Questions to check understanding

• Decide whether a given equation is of this shape, rearranging first if needed
• Compute and simplify the integrating factor for a given equation — the form of Exercise 9.5 questions 18 and 19
• Solve a given equation by this method — the form of questions 1 to 12
• Find a particular solution given a data point — the form of questions 13 to 15
• Turn a sentence about a slope into an equation of this shape and solve it — the form of questions 16 and 17
• Explain in one sentence why the left side becomes a single derivative
• Say why the constant of integration may be omitted when the factor is found
• Given an equation with y appearing squared, say why this method does not apply

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 definition (§9.4.3, Part II p. 322). A first derivative, plus a coefficient times the unknown, equal to a second coefficient — with both coefficients either constants or functions of x and nothing else. The chapter names the class and never says what "linear" means structurally. It never observes that the unknown and its derivative each occur to the first power and never inside another function, which is the property the name records and the property that makes the method available. Supply that in section 1; the Key terms table flags it as not in the book.
• The three printed instances (Part II p. 322). A constant coefficient with a sine on the right; a reciprocal of x with an exponential on the right; and a reciprocal of x times a logarithm, with a reciprocal on the right. Verified against the definition: all three qualify, and the third is worth pausing on because its coefficient looks alarming and is nonetheless a function of x alone. Add a non-instance of not in the book to the row — anything with y squared, or y inside a function — because the chapter offers none and a student needs to see the boundary, not only the interior.
• The derivation (Part II pp. 322–323). Multiply the whole equation by an unnamed function of x. Demand that the left side become the derivative of the unknown times that function. Expand with the product rule; two terms cancel; what survives is that the coefficient equals the function's derivative over the function itself. Integrate both sides and exponentiate. Verified line by line: the derivation is sound and the cancellation is exact.
• The constant the derivation drops (Part II p. 323). The integration step is written with no constant of integration and no modulus around the logarithm. Both omissions are harmless and neither is defended: a constant would multiply the factor by a fixed nonzero number, which cancels out of the final line, and the modulus can be absorbed the same way. The chapter does not say this, and a student who has just been drilled on writing the constant every time will read the omission as a mistake. Thirty seconds in section 5.
• The factor named (Part II p. 323). The chapter states that once the multiplication is done the left side has become a derivative — of a function in both letters, which it does not identify further — and gives the multiplier its name in italics with an abbreviation in brackets. That abbreviation then does all the work for the next six pages.
• The three-step recipe (Part II p. 324). Printed under a bold heading that carries no section number and is therefore invisible to any sweep looking for one. Step one: get the equation into the standard shape. Step two: compute the factor as an exponential of the integral of the first coefficient. Step three: write the unknown times the factor equal to the integral of the second coefficient times the factor, plus a constant. Step three is the safe form. — see section 11.
• Example 14 (Part II pp. 324–325). A first derivative minus the unknown, equal to a cosine. Verified: the first coefficient is minus one, so the factor is the exponential of minus x; the integral on the right returns to itself after two integrations by parts and is solved by treating it as an unknown; and the answer is half the difference of a sine and a cosine, plus a constant times the exponential of x. Note that the constant here is multiplied by the exponential — remember that when section 11 arrives.
• Example 15 (Part II p. 325). x times the derivative, plus twice the unknown, equal to x squared, with x nonzero. Verified: dividing by x first is mandatory, because the shape requires a bare derivative; then the first coefficient is twice the reciprocal of x, and the factor is the exponential of twice a logarithm, which the chapter simplifies to x squared using a bracketed identity it prints inline. The answer is a quarter of x squared plus a constant over x squared. The simplification of the factor is the teachable moment, not the integration: a factor left as an exponential of a logarithm is unusable.
• Example 17 (Part II pp. 326–327). A first derivative plus the unknown times a cotangent, equal to a sum of two terms, with a data point at a quarter turn. Verified: the factor is the exponential of the integral of the cotangent, which simplifies to a sine; the right side then splits into two integrals that cancel each other's by-parts terms exactly, leaving x squared times a sine plus a constant; and the data point makes the constant minus a quarter of pi squared. The answer is x squared minus that quantity over a sine. The cancellation is the whole difficulty and it is not signposted — the chapter simply performs it. Slow the explanation down there.
• Example 18 (Part II pp. 327–328). No equation is handed over; a sentence says the slope equals the x coordinate added to the product of the two coordinates, and asks for the curve through a stated point. Verified: the translation gives a first derivative minus x times the unknown equal to x, the factor is the exponential of minus half x squared, the integral on the right is a plain substitution, and the answer is minus one plus twice the exponential of half x squared. This example prints the two coordinate names in brackets — the only place in the chapter where either word appears — and a student who does not know them cannot read the question.
• Exercise 9.5, questions 1 to 12 (Part II pp. 328–329). Twelve general solutions. Verified, working added here on each: items one and two have constant first coefficients and are the right pair to open with; item three gives a factor of x and a cubic on the right; item four needs the standard integral of a secant and gives a factor that is a sum of a secant and a tangent; item five needs dividing by a squared cosine first and then a substitution inside the integral; item six needs dividing by x and then integration by parts on a logarithm; item seven has the reciprocal of x times a logarithm as its first coefficient and gives a logarithm as the factor; item eight needs dividing by one plus x squared and then gives the log of a sine; item nine has two terms in its first coefficient and gives a factor of x times a sine. Items ten, eleven and twelve are not of this shape at all — they are dx-over-dy items and belong to m03-t04.
• Exercise 9.5, questions 13 to 17 (Part II p. 329). Three particular solutions and two curve problems. Verified: item thirteen gives a factor that is a squared secant and an answer combining a cosine and a squared cosine; item fourteen has the factor sitting in front of it already and gives an inverse tangent minus a quarter of pi; item fifteen gives a factor that is a reciprocal cube of a sine; item sixteen is a slope condition through the origin and gives an exponential minus a linear expression; item seventeen is a slope condition through a point on the y-axis. Item seventeen is ambiguous as printed — see Notes.
• Exercise 9.5 questions 18 and 19 (Part II p. 329), both multiple choice and both asking only for the factor, not the solution. Verified: item eighteen needs dividing by x first, after which the first coefficient is minus the reciprocal of x and the factor is the reciprocal of x, the third option. Item nineteen is a dx-over-dy item and belongs to m03-t04; its answer is the reciprocal of the square root of one minus y squared, the fourth option. Two items that stop at the factor is a strong signal about what is examinable.
• The Summary bullet for this method (Part II p. 336). One bullet, and it is the thinnest in the Summary: it states the standard shape and the condition on the two coefficients, and stops. It does not name the integrating factor, it does not give the solution formula, and it does not mention the second form of the shape that §9.4.3 spends half a page on. A student revising only from the Summary can recognise one of these equations and has been told nothing about how to solve it.

Figures to have open

• A folding movement for section 3: two terms on the left of an equation shown closing into a single derivative of a product, then unfolding again. This is the load-bearing image of the topic — the whole method is that fold, and the chapter conveys it in one sentence with no picture.
• A boxed-shape card for section 1 with the two coefficient positions marked and the constraint written under each. The shape is the chapter's; the boxing is added here.
• A side-by-side constant-position frame for section 11, showing the printed displayed formula against the printed recipe step, with the arbitrary constant ringed in each and the two positions shown to disagree. Follow it with Example 14's own answer, in which the constant is multiplied by the factor's reciprocal, as the tie-breaker.
• A four-band table for section 12, built with the repo's DataTable component. Content is the shape of Exercise 9.5 as printed on Part II pp. 328–329; 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", §9.4.3 — the definition and the three printed instances, p. 322
• The derivation of the multiplier and the naming of the factor, Part II pp. 322–323
• The displayed general solution at the foot of Part II p. 323
• The three-step recipe under its unnumbered heading, Part II p. 324
• Examples 14 and 15, Part II pp. 324–325
• Example 17 with its data point, Part II pp. 326–327
• Example 18, the slope condition in words, Part II pp. 327–328
• Exercise 9.5, questions 1 to 19, Part II pp. 328–329
• Summary, the linear-equation bullet, Part II p. 336