Online calculator: Midpoint method

You can use this calculator to solve a first-degree differential equation with a given initial value using explicit midpoint method AKA modified Euler method.

To use this method, you should have differential equation in the form
$y \prime = f(x,y)$
and enter the right side of the equation f(x,y) in the y' field below.

You also need initial value as
$y(x_0)=y_0$
and the point $x$ for which you want to approximate the $y$ value.

The last parameter of a method - a step size- is a step along the tangent line to compute a function curve's successive approximation.

If you know the exact solution of a differential equation in the form y=f(x), you can enter it as well. In this case, the calculator also plots the solution along with approximation on the graph and computes the absolute error for each approximation step.

Method explanation can be found below the calculator.

Midpoint method

Initial x

Initial y

Point of approximation

Step size

Exact solution (optional)

Calculation precision

Digits after the decimal point: 2

Differential equation

Approximate value of y

Approximation

The file is very large. Browser slowdown may occur during loading and creation.

Midpoint method

As with the Euler method we use the relation
$y_{i+1}=y_i + f \Delta x$

but compute f differently. Instead of using the tangent line at the current point to advance to the next point, we are using the tangent line at the midpoint, that is, an approximate value of the derivative at the midpoint between current and next points. To do this, we approximate the y value at the midpoint as
$y_n+\frac{\Delta x}{2}f(x_n, y_n)$

And our relation changes from
$y_{i+1}=y_i + f(x_i,y_i) \Delta x$

$y_{i+1}=y_i + f(x_i+\frac{\Delta x}{2}, y_i+\frac{\Delta x}{2}f(x_i, y_i)) \Delta x$

The local error at each step of the midpoint method is of order $O\left(h^3\right)$ , giving a global error of order $O\left(h^2\right)$ . Thus, while more computationally intensive than Euler's method, the midpoint method's error generally decreases faster as $h \to 0$ .¹