# Problems Involving Maclaurin or Taylor Series Solutions of Differential Equations

Some problems at advanced Further Pure Maths level involve finding solutions to differential equations using Maclaurin or Taylor series. The choice of series depends on the given boundary conditions. I give examples below together with suggested solutions.

A Maclaurin Series Problem

Suggested solution to part (a)

Differentiating the given equation with respect to $x$ we get

$(1 + 2x) \frac{d^2 y}{d x^2} + 2 \frac{d y}{d x} = 1 + 2 \cdot 4y \frac{d y}{d x}$

$\Longleftrightarrow$

$(1 + 2x) \frac{d^2 y}{d x^2} = 1 + 2(4y - 1) \frac{d y}{d x}$

Suggested solution to part (b)

Differentiating (1) we get

$(1 + 2x) \frac{d^3 y}{d x^3} + 2 \frac{d^2 y}{d x^2} = 2(4y - 1) \frac{d^2 y}{d x^2} + 8\big(\frac{d y}{d x}\big)^2$  (Answer)

Suggested solution to part (c)

We use the Maclaurin series

$f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0)$

$\Longleftrightarrow$

$y = y_0 + x\big(\frac{d y}{d x}\big)_0 + \frac{x^2}{2!} \big(\frac{d^2 y}{d x^2}\big)_0 + \frac{x^3}{3!} \big(\frac{d^3 y}{d x^3}\big)_0$

We know that $y_0 = \frac{1}{2}$. From the first equation in the question we get upon setting $x = 0$ that

$\big(\frac{d y}{d x}\big)_0 = 4y_0^2 = 1$

From the equation in part (a) we get upon setting $x = 0$ that

$\big(\frac{d^2 y}{d x^2}\big)_0 = 1 + 2(4y_0 - 1) \cdot 1 = 3$

Finally from the equation in part (b) we get upon setting $x = 0$ that

$\big(\frac{d^3 y}{d x^3}\big)_0 + 2 \cdot 3 = 2 \cdot 3 + 8$

$\Longrightarrow$

$\big(\frac{d^3 y}{d x^3}\big)_0 = 8$

Therefore the solution is

$y \approx \frac{1}{2} + x + \frac{3x^2}{2} + \frac{4x^3}{3}$   (Answer)

A Taylor Series Problem

Suggested solution to part (a)

Differentiating the given equation with respect to $x$ we get

$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = 3 + 2 y \frac{d y}{d x}$

$\Longleftrightarrow$

$x \frac{d^2 y}{d x^2} + (1 - 2y) \frac{d y}{d x} = 3$

Suggested solution to part (b)

We use the Taylor series

$f(x) = f(1) + (x - 1)f'(1) + \frac{(x - 1)^2}{2!}f''(1) + \frac{(x - 1)^3}{3!}f'''(1)$

$\Longleftrightarrow$

$y = y_1 + (x - 1)\big(\frac{d y}{d x}\big)_1 + \frac{(x - 1)^2}{2!} \big(\frac{d^2 y}{d x^2}\big)_1 + \frac{(x - 1)^3}{3!} \big(\frac{d^3 y}{d x^3}\big)_1$

We know that $y_1 = 1$. Using this in the first equation in the question we get upon setting $x = 1$ that

$\big(\frac{d y}{d x}\big)_1 = 4$

From the second equation in the question we get upon setting $x = 1$ that

$\big(\frac{d^2 y}{d x^2}\big)_1 + (1 - 2) \cdot 4 = 3$

$\Longrightarrow$

$\big(\frac{d^2 y}{d x^2}\big)_1 = 7$

To find $\big(\frac{d^3 y}{d x^3}\big)_1$ we differentiate the second equation in the question and then set $x = 1$ to get

$\big(\frac{d^3 y}{d x^3}\big)_1 + 2 \big(\frac{d^2 y}{d x^2}\big)_1(1 - y_1) - 2 \big(\frac{d y}{d x}\big)_1^2 = 0$

$\Longrightarrow$

$\big(\frac{d^3 y}{d x^3}\big)_1 = 32$

Therefore the required series solution is

$y \approx 1 + 4(x - 1) + \frac{7(x - 1)^2}{2} + \frac{16(x - 1)^3}{3}$  (Answer)