What is the step length h?

The constant gap between successive x values, xn=x0+nh.

Why does the central difference method need two starting values?

Each step uses y{n-1}, so both y0 and y1 are needed.

How is y1 found for the central difference method?

By the forward difference method: y1=y0+hf(x0,y0).

How do you find y1 for a second-order equation?

Combine the central difference at n=0 with the recurrence formula at n=0, to eliminate y{-1}.

How is the percentage error calculated?

\frac{exact-estimate}{exact}\times100.

How can a numerical estimate be improved?

Use a smaller step length, so the difference approximations are more accurate.

What must be done when the equation contains both \frac{dy}{dx} and \frac{d^2y}{dx^2}?

Replace both by their difference approximations and collect the y{n+1}, yn and y{n-1} terms.

Numerical solution of differential equationsEdexcel A-Level Further Maths: Flashcards

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State the forward difference approximation to $\frac{dy}{dx}$ at $x_n$.

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State the forward difference approximation to $\frac{dy}{dx}$ at $x_n$ .: $\frac{y_{n+1}-y_n}{h}$ .
State the central difference approximation to $\frac{dy}{dx}$ at $x_n$ .: $\frac{y_{n+1}-y_{n-1}}{2h}$ .
State the approximation to $\frac{d^2y}{dx^2}$ at $x_n$ .: $\frac{y_{n+1}-2y_n+y_{n-1}}{h^2}$ .
Give the formula for $y_{n+1}$ from the forward difference method.: $y_{n+1}=y_n+hf(x_n,y_n)$ .
Give the formula for $y_{n+1}$ from the central difference method.: $y_{n+1}=y_{n-1}+2hf(x_n,y_n)$ .
Give the formula for $y_{n+1}$ for $\frac{d^2y}{dx^2}=f(x,y)$ .: $y_{n+1}=2y_n-y_{n-1}+h^2f(x_n,y_n)$ .
What is the step length $h$ ?: The constant gap between successive $x$ values, $x_n=x_0+nh$ .
Why does the central difference method need two starting values?: Each step uses $y_{n-1}$ , so both $y_0$ and $y_1$ are needed.
How is $y_1$ found for the central difference method?: By the forward difference method: $y_1=y_0+hf(x_0,y_0)$ .
What do the initial conditions give for a second-order equation?: $y_0$ and the gradient at $x_0$ .
How do you find $y_1$ for a second-order equation?: Combine the central difference at $n=0$ with the recurrence formula at $n=0$ , to eliminate $y_{-1}$ .
How is the percentage error calculated?: $\frac{\text{exact}-\text{estimate}}{\text{exact}}\times100$ .
How can a numerical estimate be improved?: Use a smaller step length, so the difference approximations are more accurate.
What must be done when the equation contains both $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ ?: Replace both by their difference approximations and collect the $y_{n+1}$ , $y_n$ and $y_{n-1}$ terms.

Exam questions on Numerical solution of differential equations

The differential equation $\frac{dy}{dx}=x+y$ , with $y=1$ when $x=0$ , is solved numerically using the approximation $\left(\frac{dy}{dx}\right)_n=\frac{y_{n+1}-y_n}{h}$ with $h=0.2$ . Here $x_n=0.2n$ and $y_n$ is the approximation to $y$ at $x_n$ .
Given that $y_3=1.856$ , and that the exact solution is $y=2e^x-x-1$ , calculate the percentage error in $y_3$ as an approximation to $y(0.6)$ .2 marks
The differential equation $\frac{dy}{dx}=x-y$ , with $y=1$ when $x=0$ , is solved numerically using the approximation $\left(\frac{dy}{dx}\right)_n=\frac{y_{n+1}-y_{n-1}}{2h}$ with $h=0.1$ . Here $x_n=0.1n$ , $y_0=1$ , and the value $y_1=0.9$ is found from the approximation $\left(\frac{dy}{dx}\right)_n=\frac{y_{n+1}-y_n}{h}$ .
Find $y_3$ .2 marks
The function $y$ satisfies $\frac{d^2y}{dx^2}=-y$ , with $y=0$ and $\frac{dy}{dx}=1$ when $x=0$ . It is solved numerically with step length $h=0.1$ , where $x_n=0.1n$ and $y_n$ approximates $y$ at $x_n$ , using $\left(\frac{d^2y}{dx^2}\right)_n=\frac{y_{n+1}-2y_n+y_{n-1}}{h^2}$ and $\left(\frac{dy}{dx}\right)_n=\frac{y_{n+1}-y_{n-1}}{2h}$ .
Show that $y_{n+1}=1.99y_n-y_{n-1}$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).