---
id: 59880443fb36441083c6c20e
title: Euler method
challengeType: 5
forumTopicId: 302258
dashedName: euler-method
---

# --description--

Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in [the wikipedia page](https://en.wikipedia.org/wiki/Euler method "wp: Euler method").

The ODE has to be provided in the following form:

<ul style='list-style: none;'>
  <li><big>$\frac{dy(t)}{dt} = f(t,y(t))$</big></li>
</ul>

with an initial value

<ul style='list-style: none;'>
  <li><big>$y(t_0) = y_0$</big></li>
</ul>

To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:

<ul style='list-style: none;'>
  <li><big>$\frac{dy(t)}{dt}  \approx \frac{y(t+h)-y(t)}{h}$</big></li>
</ul>

then solve for $y(t+h)$:

<ul style='list-style: none;'>
  <li><big>$y(t+h) \approx y(t) + h \, \frac{dy(t)}{dt}$</big></li>
</ul>

which is the same as

<ul style='list-style: none;'>
  <li><big>$y(t+h) \approx y(t) + h \, f(t,y(t))$</big></li>
</ul>

The iterative solution rule is then:

<ul style='list-style: none;'>
  <li><big>$y_{n+1} = y_n + h \, f(t_n, y_n)$</big></li>
</ul>

where $h$ is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.

**Example: Newton's Cooling Law**

Newton's cooling law describes how an object of initial temperature $T(t_0) = T_0$ cools down in an environment of temperature $T_R$:

<ul style='list-style: none;'>
  <li><big>$\frac{dT(t)}{dt} = -k \, \Delta T$</big></li>
</ul>

or

<ul style='list-style: none;'>
  <li><big>$\frac{dT(t)}{dt} = -k \, (T(t) - T_R)$</big></li>
</ul>

It says that the cooling rate $\\frac{dT(t)}{dt}$ of the object is proportional to the current temperature difference $\\Delta T = (T(t) - T_R)$ to the surrounding environment.

The analytical solution, which we will compare to the numerical approximation, is

<ul style='list-style: none;'>
  <li><big>$T(t) = T_R + (T_0 - T_R) \; e^{-k t}$</big></li>
</ul>

# --instructions--

Implement a routine of Euler's method and then use it to solve the given example of Newton's cooling law for three different step sizes of:

<ul>
  <li><code>2 s</code></li>
  <li><code>5 s</code> and</li>
  <li><code>10 s</code></li>
</ul>

and compare with the analytical solution.

**Initial values:**

<ul>
  <li>initial temperature <big>$T_0$</big> shall be <code>100 °C</code></li>
  <li>room temperature <big>$T_R$</big> shall be <code>20 °C</code></li>
  <li>cooling constant <big>$k$</big> shall be <code>0.07</code></li>
  <li>time interval to calculate shall be from <code>0 s</code> to <code>100 s</code></li>
</ul>

First parameter to the function is initial time, second parameter is initial temperature, third parameter is elapsed time and fourth parameter is step size.

# --hints--

`eulersMethod` should be a function.

```js
assert(typeof eulersMethod === 'function');
```

`eulersMethod(0, 100, 100, 2)` should return a number.

```js
assert(typeof eulersMethod(0, 100, 100, 2) === 'number');
```

`eulersMethod(0, 100, 100, 2)` should return 20.0424631833732.

```js
assert.equal(eulersMethod(0, 100, 100, 2), 20.0424631833732);
```

`eulersMethod(0, 100, 100, 5)` should return 20.01449963666907.

```js
assert.equal(eulersMethod(0, 100, 100, 5), 20.01449963666907);
```

`eulersMethod(0, 100, 100, 10)` should return 20.000472392.

```js
assert.equal(eulersMethod(0, 100, 100, 10), 20.000472392);
```

# --seed--

## --seed-contents--

```js
function eulersMethod(x1, y1, x2, h) {

}
```

# --solutions--

```js
function eulersMethod(x1, y1, x2, h) {
  let x = x1;
  let y = y1;

  while ((x < x2 && x1 < x2) || (x > x2 && x1 > x2)) {
    y += h * (-0.07 * (y - 20));
    x += h;
  }

  return y;
}
```
