Maths › Further Pure 1 › Series solutions of differential equations
Series solutions of differential equations
When a differential equation resists 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 x = 0 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.
WHAT YOU PROBABLY THINK
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 out. Differentiate the whole equation and substitute again: the third derivative appears. Each round yields the next coefficient of the solution's Taylor series, so the equation itself refutes the opening claim, one derivative at a time.
WORKED EXAMPLE
The specification's own equation
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, which is 3.
Taylor: y = 1 − x²/2! + 3x⁴/4! = 1 − x²/2 + x⁴/8 + …
Spot check at x = 0.4: the series gives 0.9232 and a fine numerical solution 0.9231. Four terms, four decimal places.
Reduction by substitution
Some equations are a standard type in different variables. The exam supplies the substitution; the skill is to transform both the derivative and the function, then solve the familiar equation and translate 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: 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.
YOUR TURN
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.
THE EXAM BIT
- Rearrange for the highest derivative first; every later step reuses that arrangement.
- Differentiate the equation as it stands, products and all, before substituting numbers.
- Keep factorials visible until the final line: 3/4! reads as method, 1/8 reads as luck.
- Under a substitution, transform dy/dx explicitly; quoting the new equation unearned loses the 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
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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.
CHECK YOUR PROGRESS
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- Extract successive derivative values at x = 0 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.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.