IB Maths AA HLTopic 5 — CalculusPaper 1 & 2HL~12 min read
Maclaurin Series from Differential Equations
Given a DE dy/dx = g(x, y) with an initial condition y(0), you can build the Maclaurin series for the solution y = f(x) term by term — even when the DE has no closed-form solution. The trick: implicit differentiation gives y″, y‴, … at x = 0, one at a time. Plug into the general Maclaurin formula. Each new derivative gives the next term of the series.
📘 What you need to know
Goal: given dy/dx = g(x, y) and y(0) = a, find the Maclaurin series for y = f(x) without solving the DE explicitly.
The build-up: y(0) comes from the initial condition; y′(0) comes from the DE; y″(0), y‴(0), … come from differentiating the DE implicitly.
Maclaurin formula: y = y(0) + y′(0)·x + (y″(0)/2!)x² + (y‴(0)/3!)x³ + … (same general formula, applied to the solution function).
Implicit differentiation: y is treated as a function of x, so the chain rule gives d/dx(y²) = 2yy′; d/dx(yy′) = y′·y′ + y·y″ = (y′)² + yy″.
Use the “prime” notation: y′ for dy/dx, y″ for d²y/dx², etc. — much faster to write than Leibniz form.
To find the term in xn, you need derivatives up to y(n)(0). For a series up to x⁴, find y, y′, y″, y‴, y⁽⁴⁾ at 0.
Second-order DEs work the same way — they give you y″ directly, plus two initial conditions y(0) and y′(0); higher derivatives come by differentiating the DE.
Application: once you have the series, substitute small x values to approximate y(x) numerically — useful when the DE has no elementary solution.
The build-up: deriving each y(n)(0) in turn
Maclaurin series of the solutiony(x) = y(0) + xy′(0) + x²2!y″(0) + x³3!y‴(0) + …
Each derivative y(n) is obtained by implicitly differentiating the previous one. Evaluated at x = 0 (using the initial condition y = 1), each value feeds into the next row. The Maclaurin formula then assembles them.
Why this works even without a closed-form solution
DE with closed form
y′ = y²
Solution: y = 1/(1−x) (by separation). Maclaurin series can be derived directly from the solution, OR from the DE — both methods give the geometric series.
DE with no closed form
y′ = y² + x²
No elementary solution exists. But we can still build the Maclaurin series term by term — the DE-implicit method works even when no formula for y(x) is available.
The power of the method: this is one of only two ways to get information about solutions to “unsolvable” DEs. The other is Euler’s method (numerical). Maclaurin gives an algebraic answer; Euler gives a numerical one.
🧭 Recipe — building a Maclaurin series from a DE
Read off y(0) from the initial condition. This becomes the constant term of the Maclaurin series.
Compute y′(0) by substituting x = 0 and the known y(0) into the DE.
Differentiate the DE implicitly to get an expression for y″ in terms of x, y, and y′. Evaluate at x = 0 using known values.
Repeat: differentiate again to get y‴, then y⁽⁴⁾, etc. — as many as needed for the required series length. Substitute previously-computed values each time.
Plug into the Maclaurin formula: y = y(0) + xy′(0) + (x²/2!) y″(0) + (x³/3!) y‴(0) + … and simplify each coefficient.
Worked examples
WE 1
First-order DE — find Maclaurin up to x⁴
For the DE y′ = x + y with y(0) = 1, find the Maclaurin series of y up to and including the term in x⁴.
Step 1 — read off y(0) and find y'(0)y(0) = 1 (from IC)y'(0) = x + y at x=0, y=1 = 0 + 1 = 1Step 2 — differentiate the DE implicitlyy’ = x + y ⟹ y” = 1 + y’y”(0) = 1 + 1 = 2Step 3 — differentiate againy”’ = 0 + y” = y”y”'(0) = 2y⁽⁴⁾ = y”’ = y” (chain continues)y⁽⁴⁾(0) = 2Step 4 — plug into Maclaurin formulay = 1 + 1·x + 2·x²/2! + 2·x³/3! + 2·x⁴/4! = 1 + x + x² + x³/3 + x⁴/12y ≈ 1 + x + x² + x³3 + x⁴12cross-check: this DE has the exact solution y = −x − 1 + 2e^x. Taking its Maclaurin series: −x − 1 + 2(1 + x + x²/2 + x³/6 + x⁴/24 + …) = 1 + x + x² + x³/3 + x⁴/12 + … ✓ matches our DE-based derivation.
WE 2
A surprising result — y’ = y², y(0) = 1
Find the Maclaurin series for the solution of y′ = y² with y(0) = 1, up to and including the term in x⁴.
Step 1 — IC and y'(0)y(0) = 1y'(0) = y² at y=1 = 1Step 2 — differentiate y’ = y² implicitlyy” = 2y·y’y”(0) = 2(1)(1) = 2Step 3 — keep differentiatingy”’ = 2(y’)² + 2y·y” (product rule)y”'(0) = 2(1)² + 2(1)(2) = 2 + 4 = 6y⁽⁴⁾ = 4y’·y” + 2y’·y” + 2y·y”’ = 6y’·y” + 2y·y”’y⁽⁴⁾(0) = 6(1)(2) + 2(1)(6) = 12 + 12 = 24Step 4 — assemble the Maclaurin seriesy = 1 + 1·x + 2x²/2! + 6x³/3! + 24x⁴/4! = 1 + x + x² + x³ + x⁴y ≈ 1 + x + x² + x³ + x⁴ (the geometric series!)notice the beautiful cancellation: y⁽ⁿ⁾(0) = n! exactly, so dividing by n! gives coefficient 1 for every power. This is the geometric series, the Maclaurin expansion of y = 1/(1−x). The DE y’ = y² with y(0) = 1 is solved by 1/(1−x) (by separation), so the answer is consistent.
WE 3
Derive the tan x Maclaurin series — y’ = 1 + y², y(0) = 0
The function y = tan x satisfies the DE y′ = 1 + y² with y(0) = 0. Use this DE to find the Maclaurin series for tan x up to and including the term in x⁵.
Step 1 — IC and y'(0)y(0) = 0, y'(0) = 1 + 0² = 1Step 2 — differentiate implicitlyy” = 2y·y’y”(0) = 2(0)(1) = 0Step 3 — keep goingy”’ = 2(y’)² + 2y·y”y”'(0) = 2(1)² + 2(0)(0) = 2y⁽⁴⁾ = 6y’·y” + 2y·y”’y⁽⁴⁾(0) = 6(1)(0) + 2(0)(2) = 0y⁽⁵⁾ = 6(y”)² + 6y’·y”’ + 2y’·y”’ + 2y·y⁽⁴⁾ = 6(y”)² + 8y’·y”’ + 2y·y⁽⁴⁾y⁽⁵⁾(0) = 6(0)² + 8(1)(2) + 2(0)(0) = 16Step 4 — assembley = 0 + 1·x + 0·x²/2 + 2·x³/6 + 0·x⁴/24 + 16·x⁵/120 = x + x³/3 + 2x⁵/15tan x ≈ x + x³3 + 2x⁵15tan x is not in the formula booklet, so the DE-implicit method is a clean way to derive its series. Matches our first-principles derivation in the “Maclaurin Standard Functions” note. The even-power coefficients vanish because tan is an odd function.
WE 4
Second-order DE — y” = -y, y(0) = 1, y'(0) = 0
Use the second-order DE y″ = −y with initial conditions y(0) = 1 and y′(0) = 0 to derive the Maclaurin series of y up to and including the term in x⁶.
Step 1 — IC valuesy(0) = 1, y'(0) = 0, y”(0) = -y(0) = -1Step 2 — differentiate the DEy”’ = -y’ ⟹ y”'(0) = -y'(0) = 0y⁽⁴⁾ = -y” ⟹ y⁽⁴⁾(0) = -(-1) = 1y⁽⁵⁾ = -y”’ ⟹ y⁽⁵⁾(0) = -0 = 0y⁽⁶⁾ = -y⁽⁴⁾ ⟹ y⁽⁶⁾(0) = -1Step 3 — assemble the seriesy = 1 + 0·x + (-1)x²/2! + 0·x³/3! + 1·x⁴/4! + 0·x⁵/5! + (-1)x⁶/6! = 1 – x²/2! + x⁴/4! – x⁶/6!y ≈ 1 − x²2! + x⁴4! − x⁶6! (this is cos x!)a second-order DE provides y″ directly and supplies TWO initial conditions (y(0) and y'(0)). The pattern y”=-y has a 4-step derivative cycle: y, y’, -y, -y’, y, y’… so every fourth derivative repeats. The solution is cos x ✓.
WE 5
Numerical application — approximate y(0.1) for WE 1’s DE
The DE y′ = x + y, y(0) = 1 has Maclaurin series y = 1 + x + x² + x³/3 + x⁴/12 + … (from WE 1). Use this to approximate y(0.1) to 5 decimal places.
Step 1 — substitute x = 0.1 into the truncated seriesy(0.1) ≈ 1 + 0.1 + (0.1)² + (0.1)³/3 + (0.1)⁴/12 = 1 + 0.1 + 0.01 + 0.001/3 + 0.0001/12Step 2 — compute each term1 + 0.1 = 1.10.01 → running total 1.110.001/3 ≈ 0.000333 → running total ≈ 1.1103330.0001/12 ≈ 0.00000833 → total ≈ 1.110342y(0.1) ≈ 1.11034 (to 5 d.p.)exact value: y(0.1) = −0.1 − 1 + 2e^0.1 = 1.110342… Our 5-term Maclaurin approximation matches to about 7 decimal places — very accurate near x = 0. The smaller |x| is, the fewer terms you need.
WE 6
A DE with no closed-form solution — y’ = y² + x²
The DE y′ = y² + x² with y(0) = 1 has no elementary closed-form solution. Find the Maclaurin series of its solution up to and including the term in x⁴, then approximate y(0.2).
Step 1 — IC and derivatives at x = 0y(0) = 1y'(0) = y² + x² at x=0, y=1 = 1 + 0 = 1Step 2 — implicit differentiationy’ = y² + x²y” = 2yy’ + 2xy”(0) = 2(1)(1) + 0 = 2Step 3 — keep differentiatingy”’ = 2(y’)² + 2yy” + 2y”'(0) = 2(1)² + 2(1)(2) + 2 = 2 + 4 + 2 = 8y⁽⁴⁾ = 6y’y” + 2yy”’ (the +2 differentiates to 0)y⁽⁴⁾(0) = 6(1)(2) + 2(1)(8) = 12 + 16 = 28Step 4 — assemble Maclaurin seriesy = 1 + 1·x + 2x²/2 + 8x³/6 + 28x⁴/24 = 1 + x + x² + 4x³/3 + 7x⁴/6Step 5 — approximate y(0.2)y(0.2) ≈ 1 + 0.2 + 0.04 + (4/3)(0.008) + (7/6)(0.0016) ≈ 1 + 0.2 + 0.04 + 0.01067 + 0.00187 ≈ 1.25253y(x) ≈ 1 + x + x² + 4x³3 + 7x⁴6; y(0.2) ≈ 1.25253no closed-form solution exists for this DE — yet we got a useful approximation. This is one of two ways to attack “unsolvable” DEs (the other being Euler’s method, which gives numerical estimates without an algebraic formula).
💡 Top tips
Use prime notation: y′, y″, y‴, y⁽⁴⁾. Much faster to write than dy/dx, d²y/dx², etc. — and clearer on the page.
For xn, you need y(n)(0). Plan ahead: if the question asks for terms up to x⁴, you’ll compute up to y⁽⁴⁾(0).
Substitute as you go: at each step, plug in the previously-computed y(0), y′(0), etc. — don’t carry symbolic expressions further than necessary.
Watch for cancellations / zeros: if y″(0) = 0, that doesn’t mean you can stop — the x² term just vanishes, but you still need y‴(0), y⁽⁴⁾(0) etc.
Cross-check when possible: if a closed-form solution is easy to find (by separation, IF, etc.), verify the first few coefficients of your DE-derived series match its Maclaurin expansion.
⚠ Common mistakes
Forgetting the chain rule: d/dx(y²) is 2y·y′ (with the implicit y′), NOT 2y. Same for sin y, ey, etc.
Product-rule slips: d/dx(y·y′) = (y′)² + y·y″, NOT just y·y″. Both factors are functions of x via implicit differentiation.
Forgetting the factorial denominators: coefficient of xn in Maclaurin is y(n)(0)/n!, not just y(n)(0).
Computing y(n)(0) too early: you usually need to keep the symbolic expression in terms of y, y′, y″, … until just before substituting x = 0. Don’t lose information.
Stopping at the wrong order: “up to x⁴” means terms in x⁴ are included. You need y⁽⁴⁾(0) to compute that coefficient.
You’ve finished the Maclaurin Series chapter! 🎉 You now know four routes to a Maclaurin series: first-principles (derivatives at 0), substitution/multiplication (composites & products), differentiation/integration of known series, and DE-implicit (when only a DE is given). Up next: l’Hôpital’s Rule & Maclaurin for Limits — using these series to crack indeterminate-form limits that ordinary algebra can’t touch.
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