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Worked solution

Deduce a related series from an implicit differential equation

The curve y=f(x)y=f(x) passes through (0,π/4)(0,\pi/4) and satisfies y′tan⁡y=ln⁡((ex+1)/2)y^\prime\tan y=\ln((e^x+1)/2). Find the first three nonzero Maclaurin terms of yy. Deduce the expansion of (cot⁡y)ln⁡((ex+1)/2)(\cot y)\ln((e^x+1)/2) through x2x^2.

[2007/YJC/II/1]

CJC · Discussion · Question 9

Answer

y=π/4+x2/4+x3/24+O(x4)y=\pi/4+x^2/4+x^3/24+O(x^4); the deduced expression is x/2+x2/8+O(x3)x/2+x^2/8+O(x^3).

Differentiate the relation

Write the right side as ln⁡(ex+1)−ln⁡2\ln(e^x+1)-\ln2. Differentiating gives sec⁡2y (y′)2+tan⁡y y′′=ex/(ex+1)\sec^2y\,(y^\prime)^2+\tan y\,y^{\prime\prime}=e^x/(e^x+1). A second differentiation gives 2sec⁡2ytan⁡y (y′)3+3sec⁡2y y′y′′+tan⁡y y′′′=ex/(ex+1)−e2x/(ex+1)22\sec^2y\tan y\,(y^\prime)^3+3\sec^2y\,y^\prime y^{\prime\prime}+\tan y\,y^{\prime\prime\prime}=e^x/(e^x+1)-e^{2x}/(e^x+1)^2.

How did this part go?

Evaluate and retain the first three nonzero terms

At zero, y=π/4y=\pi/4. The original relation gives y′(0)=0y^\prime(0)=0; the differentiated relations give y′′(0)=1/2y^{\prime\prime}(0)=1/2 and y′′′(0)=1/4y^{\prime\prime\prime}(0)=1/4. Thus y=π/4+x2/4+x3/24+O(x4)y=\pi/4+x^2/4+x^3/24+O(x^4).

How did this part go?

Recognise the requested expression as y′

The original equation gives (cot⁡y)ln⁡((ex+1)/2)=y′(\cot y)\ln((e^x+1)/2)=y^\prime. Differentiating the analytic series yields y′=x/2+x2/8+O(x3)y^\prime=x/2+x^2/8+O(x^3).

How did this part go?

Common mistakes

The zero coefficient of x does not count among the first three nonzero terms.

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