# Euler Method for solving differential equation

Given a differential equation dy/dx = f(x, y) with initial condition y(x0) = y0. Find its approximate solution using Euler method.

Euler Method :
In mathematics and computational science, the Euler method (also called forward
Euler method) is a first-order numerical procedurefor solving ordinary differential
equations (ODEs) with a given initial value.
Consider a differential equation dy/dx = f(x, y) with initialcondition y(x0)=y0
then succesive approximation of this equation can be given by:

y(n+1) = y(n) + h * f(x(n), y(n))

where h = (x(n) – x(0)) / n
h indicates step size. Choosing smaller
values of h leads to more accurate results
and more computation time.

Example :

```
Consider below differential equation
dy/dx = (x + y + xy)
with initial condition y(0) = 1
and step size h = 0.025.
Find y(0.1).

Solution:
f(x, y) = (x + y + xy)
x0 = 0, y0 = 1, h = 0.025
Now we can calculate y1 using Euler formula
y1 = y0 + h * f(x0, y0)
y1 = 1 + 0.025 *(0 + 1 + 0 * 1)
y1 = 1.025
y(0.025) = 1.025.
Similarly we can calculate y(0.050), y(0.075), ....y(0.1).
y(0.1) = 1.11167
```

## C++

 `/* CPP  Program to find approximation ` `   ``of a ordinary differential equation ` `   ``using euler method.*/` `#include ` `using` `namespace` `std; ` ` `  `// Consider a differential equation ` `// dy/dx=(x + y + xy) ` `float` `func(``float` `x, ``float` `y) ` `{ ` `    ``return` `(x + y + x * y); ` `} ` ` `  `// Function for Euler formula ` `void` `euler(``float` `x0, ``float` `y, ``float` `h, ``float` `x) ` `{ ` `    ``float` `temp = -0; ` ` `  `    ``// Iterating till the point at which we ` `    ``// need approximation ` `    ``while` `(x0 < x) { ` `        ``temp = y; ` `        ``y = y + h * func(x0, y); ` `        ``x0 = x0 + h; ` `    ``} ` ` `  `    ``// Printing approximation ` `    ``cout << ``"Approximate solution at x = "` `         ``<< x << ``"  is  "` `<< y << endl; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``// Initial Values ` `    ``float` `x0 = 0; ` `    ``float` `y0 = 1; ` `    ``float` `h = 0.025; ` ` `  `    ``// Value of x at which we need approximation ` `    ``float` `x = 0.1; ` ` `  `    ``euler(x0, y0, h, x); ` `    ``return` `0; ` `} `

## Python3

 `# Python Code to find approximation ` `# of a ordinary differential equation ` `# using euler method. ` ` `  `# Consider a differential equation ` `# dy / dx =(x + y + xy) ` `def` `func( x, y ): ` `    ``return` `(x ``+` `y ``+` `x ``*` `y) ` `     `  `# Function for euler formula ` `def` `euler( x0, y, h, x ): ` `    ``temp ``=` `-``0` ` `  `    ``# Iterating till the point at which we ` `    ``# need approximation ` `    ``while` `x0 < x: ` `        ``temp ``=` `y ` `        ``y ``=` `y ``+` `h ``*` `func(x0, y) ` `        ``x0 ``=` `x0 ``+` `h ` ` `  `    ``# Printing approximation ` `    ``print``(``"Approximate solution at x = "``, x, ``" is "``, ``"%.6f"``%` `y) ` `     `  `# Driver Code ` `# Initial Values ` `x0 ``=` `0` `y0 ``=` `1` `h ``=` `0.025` ` `  `# Value of x at which we need approximation ` `x ``=` `0.1` ` `  `euler(x0, y0, h, x) `

## Java

 `// Java program to find approximation of an ordinary ` `// differential equation using euler method ` `import` `java.io.*; ` ` `  `class` `Euler { ` `    ``// Consider a differential equation ` `    ``// dy/dx=(x + y + xy) ` `    ``float` `func(``float` `x, ``float` `y) ` `    ``{ ` `        ``return` `(x + y + x * y); ` `    ``} ` ` `  `    ``// Function for Euler formula ` `    ``void` `euler(``float` `x0, ``float` `y, ``float` `h, ``float` `x) ` `    ``{ ` `        ``float` `temp = -``0``; ` ` `  `        ``// Iterating till the point at which we ` `        ``// need approximation ` `        ``while` `(x0 < x) { ` `            ``temp = y; ` `            ``y = y + h * func(x0, y); ` `            ``x0 = x0 + h; ` `        ``} ` ` `  `        ``// Printing approximation ` `        ``System.out.println(``"Approximate solution at x = "` `                           ``+ x + ``" is "` `+ y); ` `    ``} ` ` `  `    ``// Driver program ` `    ``public` `static` `void` `main(String args[]) ``throws` `IOException ` `    ``{ ` `        ``Euler obj = ``new` `Euler(); ` `        ``// Initial Values ` `        ``float` `x0 = ``0``; ` `        ``float` `y0 = ``1``; ` `        ``float` `h = ``0``.025f; ` ` `  `        ``// Value of x at which we need approximation ` `        ``float` `x = ``0``.1f; ` ` `  `        ``obj.euler(x0, y0, h, x); ` `    ``} ` `} ` ` `  `// This code is contributed by Anshika Goyal. `

## C#

 `// C# program to find approximation of an ordinary ` `// differential equation using euler method ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Consider a differential equation ` `    ``// dy/dx=(x + y + xy) ` `    ``static` `float` `func(``float` `x, ``float` `y) ` `    ``{ ` `        ``return` `(x + y + x * y); ` `    ``} ` ` `  `    ``// Function for Euler formula ` `    ``static` `void` `euler(``float` `x0, ``float` `y, ``float` `h, ``float` `x) ` `    ``{ ` ` `  `        ``// Iterating till the point at which we ` `        ``// need approximation ` `        ``while` `(x0 < x) { ` `            ``y = y + h * func(x0, y); ` `            ``x0 = x0 + h; ` `        ``} ` ` `  `        ``// Printing approximation ` `        ``Console.WriteLine(``"Approximate solution at x = "` `                          ``+ x + ``" is "` `+ y); ` `    ``} ` ` `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{ ` ` `  `        ``// Initial Values ` `        ``float` `x0 = 0; ` `        ``float` `y0 = 1; ` `        ``float` `h = 0.025f; ` ` `  `        ``// Value of x at which we need ` `        ``// approximation ` `        ``float` `x = 0.1f; ` ` `  `        ``euler(x0, y0, h, x); ` `    ``} ` `} ` ` `  `// This code is contributed by Vt_m. `

## PHP

 ` `

Output :

```Approximate solution at x = 0.1  is  1.11167
```

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