Practise › Questions › Series solutions of differential equations
Series solutions of differential equations questions
When a differential equation has no solution in closed form, its own terms can supply a Taylor series. Repeated differentiation of the equation gives the value of each derivative at the starting point, and those values become the coefficients of the series.
6 original questions · 22 marks · the series solutions of differential equations notes · Further Pure 1
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Describe the Taylor series method for solving a differential equation with given initial conditions.
Worked answer
Rearrange for the highest derivative and substitute the initial values to find it at the starting point. Differentiate the whole equation and substitute again for the next derivative, and so on. Each derivative becomes one coefficient of the Taylor series of the solution. B1 for evaluating the highest derivative from the initial values, B1 for repeated differentiation supplying each further coefficient.For dy/dx = x + y² with y = 1 at x = 0, find the values of the first and second derivatives of y at x = 0.
Worked answer
dy/dx at 0 is 0 + 1² = 1. Differentiating the equation: the second derivative is 1 + 2y(dy/dx), which at 0 is 1 + 2(1)(1) = 3. B1 for dy/dx = 1, M1 for differentiating the equation, A1 for 3.For dy/dx = x + y² with y = 1 at x = 0, the first and second derivatives of y at x = 0 are 1 and 3. Find the third derivative at x = 0 and write the series solution up to the term in x³.
Worked answer
The second derivative is 1 + 2y y′. Differentiating that by the product rule gives 2(y′)² + 2y y″, which at the origin is 2(1)² + 2(1)(3) = 8.
Taylor's series about 0 is y = y(0) + y′(0)x + y″(0)x²/2! + y‴(0)x³/3!, so y = 1 + x + 3x²/2! + 8x³/3!, that is y = 1 + x + 1.5x² + (4/3)x³ + …. M1 for differentiating by the product rule, A1 for 8, M1 for the Taylor form with its factorials, A1 for the series. Keep the factorials visible until the last line; dividing by n rather than n! is the standard slip.Use the substitution z = y² to solve 2y(dy/dx) + y² = x.
Worked answer
dz/dx = 2y(dy/dx), so the equation becomes dz/dx + z = x: linear in z. The integrating factor is ex, giving exz = ∫x ex dx = (x − 1)ex + C. So z = x − 1 + Ce−x, and y² = x − 1 + Ce−x. B1 for dz/dx = 2y(dy/dx), M1 for the integrating factor ex, A1 for exz = (x − 1)ex + C, A1 for y² = x − 1 + Ce−x. Substituting back into the original equation confirms it.For a second order equation, explain why two initial conditions are needed before the Taylor series method can begin.
Worked answer
The equation delivers the second derivative in terms of x, y and dy/dx, so both y and dy/dx must be known at the starting point before anything can be evaluated. With only one condition the very first substitution is impossible. B1 B1 for the two points: the equation returns the second derivative in terms of y and dy/dx, and both must be known before any substitution.Given that y'' + xy' + 2y = 0 with y = 1 and dy/dx = 0 at x = 0, find the series solution for y up to and including the term in x⁶. Hence estimate the value of y at x = 0.2, giving your answer to five decimal places.
Worked answer
Rearrange for the highest derivative: y″ = −xy′ − 2y. At the origin y″ = −0 − 2(1) = −2.
Differentiate, using the product rule on xy′:
y‴ = −y′ − xy″ − 2y′ = −3y′ − xy″, so y‴(0) = 0.
Again: y⁗ = −3y″ − y″ − xy‴ = −4y″ − xy‴, so y⁗(0) = −4(−2) = 8.
The pattern holds, each differentiation raising the coefficient by one. The fifth derivative is −5y‴ − xy⁗, giving 0, and the sixth is −6y⁗ − xy⁽⁵⁾, giving −6(8) = −48.
Assembling Taylor's series:
y = 1 + 0 − 2x²/2! + 0 + 8x⁴/4! + 0 − 48x⁶/6!
= 1 − x² + x⁴/3 − x⁶/15.
Every odd derivative vanishes, because each is a multiple of the one two places before it and the chain starts at y′(0) = 0. Spotting that halves the work.
At x = 0.2: y ≈ 1 − 0.04 + 0.0016/3 − 0.000064/15 = 1 − 0.04 + 0.00053333 − 0.00000427 = 0.96053.
M1 for rearranging for y″, A1 for y″(0) = −2, M1 for the repeated differentiation, A1 for y⁗(0) = 8, A1 for −48 at the sixth derivative, A1 for the assembled series, B1 for 0.96053. The terms shrink by roughly a factor of a hundred each time, so the next one cannot disturb the fifth decimal place.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise series solutions of differential equations one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.