Maths › Further Pure 1 › Series solutions of differential equations
Series solutions of differential equations
When a differential equation resists an exact solution, make it dictate its own Taylor series. The equation hands over every derivative at the starting point, term by factorial term.
Builds on Taylor series and First order equations and integrating factors.
IN THIS TOPIC
- Extract successive derivative values at the starting point directly from the equation.
- Assemble the Taylor series of the solution as far as a stated power.
- Apply a given substitution to reduce an equation to a solvable type.
COMMON MISCONCEPTION
A differential equation you cannot solve exactly tells you nothing about its solution.
The equation as a derivative factory
Rearrange the equation for the highest derivative and substitute the initial values. The second derivative at 0 falls straight out. Now differentiate the whole equation and substitute again, and the third derivative appears. Each round yields the next coefficient of the solution's Taylor series, so the equation itself hands over the solution, one derivative at a time.
WORKED EXAMPLE
A worked series solution
Find a series solution of y'' + xy' + y = 0, with y = 1 and y' = 0 at x = 0, up to the term in x⁴.
At x = 0: y'' = −xy' − y = −1.
Differentiate. The third derivative is −2y' − xy'', which is 0 at x = 0. Once more, the fourth is −3y'' − x times the third, so it takes the value 3.
Taylor: y = 1 − x²/2! + 3x⁴/4! = 1 − x²/2 + x⁴/8 + …
This one happens to have the closed form y = e−x²/2, and its expansion agrees term for term. At x = 0.4 the series gives 0.9232 and the exact value is 0.9231.
Where the marks are won and lost
Two things go wrong in scripts, both avoidable. The first is careless differentiation of the equation. In y'' + xy' + y = 0 the middle term is a product, so differentiating it gives y' + xy'', and candidates who drop that first piece poison every coefficient after it.
The second is tidying too early. Keep the factorials on display while you build the series, because 3/4! shows the examiner a method and 1/8 shows nothing at all. Only simplify on the final line.
Nothing forces the anchor to sit at zero. If the conditions are given at x = 2, every derivative is evaluated there and the answer comes out in powers of (x − 2). The method is identical, and only the bracket changes.
Reduction by substitution
Some equations are a standard type written in different variables. The exam supplies the substitution, and the skill lies in transforming both the derivative and the function, then solving the familiar equation and translating the answer back.
WORKED EXAMPLE
A reciprocal substitution
Use z = 1/y to solve dy/dx + y = xy².
dz/dx = −(1/y²)(dy/dx), so dividing the equation by −y² gives dz/dx − z = −x, which is linear in z.
Integrating factor e−x: z = x + 1 + Cex.
So y = 1/(x + 1 + Cex). Substituting back into the original equation confirms it exactly.
GUIDED PRACTICE
Second derivative from the equation
For y'' = x + y² with y(0) = 1 and y'(0) = 2, find the second and third derivatives at 0, and write the series solution up to x³.
Show the working
y''(0) = 0 + 1² = 1.
Differentiating gives 1 + 2yy' for the third derivative, which at 0 is 1 + 2(1)(2) = 5.
y = 1 + 2x + x²/2 + 5x³/6 + … The equation manufactured both coefficients without ever being solved.
ASSESSMENT FOCUS
- Rearrange for the highest derivative first, since every later step reuses that arrangement.
- Differentiate the equation as it stands, products and all, before substituting any numbers.
- Keep the factorials visible until the final line.
- Under a substitution, transform dy/dx explicitly. Quoting the new equation unearned loses the method marks.
CHECK YOURSELF
For y' = x² + y with y(0) = 1, find the second and third derivatives of y at x = 0.
Show a hint
Differentiate the equation itself, then substitute x = 0.
Show the answer
y'' = 2x + y', and y'(0) = 1, so y''(0) = 1. Differentiating again gives 2 + y'', so the third derivative at 0 is 3.
Solve for the highest derivative, then differentiate and substitute repeatedly; each round yields the next Taylor coefficient.
A given substitution turns the equation into a standard type. Transform the derivatives, solve, translate back.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the series solutions of differential equations questions page.
CHECK YOUR PROGRESS
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- Extract successive derivative values at the starting point directly from the equation.
- Assemble the Taylor series of the solution as far as a stated power.
- Apply a given substitution to reduce an equation to a solvable type.
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