Let’s recall the very basic aspect of Differential Equation.

We know,          $\frac{\text{d}y(x)}{\text{d}x}=f(x,y)$

It can be expressed as,       $\frac{\text{d}y(x)}{\text{d}x}\sim \frac {y(x+h)-y(x)}{h}$

Combining both the equations, we get,       $\frac{\text{d}y(x)}{\text{d}x}\sim \frac {y(x+h)-y(x)}{h}=f(x,y)$

Summarizing the equation as,     $y(x+h)= y(x)=h.f(x,y)$

Thus, we arrive at,      $y_{n+1}= y_{n}=h.f(x_{n},y_{n})$.

where $h$ is the step size (the smaller, the better), $f(x,y)$ is right side of the differential equation.

Example 1:            $\frac{\text{d}y}{\text{d}x}=2y$ with the initial condition as    $y(0)=1$.

Here, we have two approaches to find the solution of the Differential Equation, which are as follows:

1) Analytical Solution: This is the exact solution of an ordinary differential equation.

Integrating both sides, $\int_{}{}\frac{\text{d}y}{ y}=\int_{}{}{d}x$

We get, $\ln y = x + \ln C$

Therefore, $y(x) = e^{x}$

2) Numerical Solution: This is just an approximation of the solution.

We require this formula, $y(x+h)= y(x)=h.f(x,y)$

along with initial conditions,     $x_{0}=0$ and $y_{0}=1$

Let’s continue with Example 1,  to find the Numerical Solution.

Example 1:            $\frac{\text{d}y}{\text{d}x}=2y$ with the initial condition as   $y(0)=1$ ad step-size $h=2$.

We know the formula,  $y_{n+1}= y_{n}=h.f(x_{n},y_{n})$.

In this equation we put the values of $x_{n}$ at interval of $2$ because of Step-Size ($h=2$) and obtain the value of $y_{n}.$ By initial conditions, we know $x_{0}=0$ and $y_{0}=1$

Here, we are assuming the slope to be same between two consecutive points of $x$ and $y$, whereas in the exact case the slope is changing constantly.