We start at x = 0 and move to x=2, with a step size of 1. Essentially, we approximate the next step by using the formula:

$\displaystyle p(x + \Delta x) = p(x) + p'(x)\cdot (\Delta x)$.

So applying Euler's method, we evaluate using derivative: 

$\displaystyle p'(x) = 0.5x(1-x), \ p(0) = 2$ 

And two step sizes, at x = 1 and x = 2.

$\displaystyle {}p(x=1) = p(0) + (1 - 0)p'(0) = 2 + 1\cdot p'(0) = 2 + 1 \cdot 0.5(0)(1-0) = 2 + 0 = 2$ 

$\displaystyle {}p(x=2) = p(1) + (2-1)\cdot p'(1) = 2 + 1\cdot p'(1) = 2 + 0.5(1)(1-1) = 2 + 0 = 2$

And thus the evaluation of p at x = 2, using Euler's method, gives us p(2) = 2.

Using Euler's method with $\displaystyle \small h=1/4$ means that we use two iterations to get the approximation. The general iterative formula is 

$\displaystyle \small y_{i+1}=y_i+h\cdot f(t_i,y_i)$

where each $\displaystyle \small t_i$ is

$\displaystyle \small t_0=0$

$\displaystyle \small t_1=\frac{1}{4}$

$\displaystyle \small t_2=\frac{1}{2}$

 $\displaystyle \small y_i$ is an approximation of $\displaystyle \small y(t_i)$, and $\displaystyle y'=f(t,y)=y$, for this differential equation. So we have

$\displaystyle \small \small \small \small y_1=y_0+\frac{1}{4}f(t_0,y_0)=1+\frac{1}{4}\cdot 1=1.25$

$\displaystyle \small \small \small \small e^{1/2}= y(\frac{1}{2})\approx y_2=y_1+\frac{1}{4}f(t_1,y_1)=1.25+\frac{1}{4}\cdot 1.25=1.5625$

So our approximation of $\displaystyle \small e^{1/2}$ is

$\displaystyle \small e^{1/2}\approx 1.5625$

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,

$\displaystyle f(x,y)=\frac{dy}{dx}=3x+4y$ 

$\displaystyle h=0.25$

Starting at the initial point $\displaystyle (x_0,y_0)=(0,0)$

$\displaystyle x_{1}=x_0+h=0+0.25=0.25$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0+0.25*(3*0+4*0)=0$

$\displaystyle (x_1,y_1)=(0.25,0)$

$\displaystyle x_{2}=x_1+h=0.25+0.25=0.5$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=0+0.25*(3*0.25+4*0)=0.1875$

$\displaystyle (x_2,y_2)=(0.5,0.1875)$

We continue using Euler's method until $\displaystyle x=1$.  The results of Euler's method are in the table below.

Note: Solving this differential equation analytically gives a different answer, $\displaystyle 9.2997$.  Future problems will explain this discrepancy. 

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,

$\displaystyle f(x,y)=\frac{dy}{dx}=3x+4y$ 

$\displaystyle h=0.1$

Starting at the initial point $\displaystyle (x_0,y_0)=(0,0)$

$\displaystyle x_{1}=x_0+h=0+0.1=0.1$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0+0.1*(3*0+4*0)=0$

$\displaystyle (x_1,y_1)=(0.1,0)$

$\displaystyle x_{2}=x_1+h=0.1+0.1=0.2$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=0+0.1*(3*0.1+4*0)=0.03$

$\displaystyle (x_2,y_2)=(0.2,0.03)$

We continue using Euler's method until $\displaystyle x=1$.  The results of Euler's method are in the table below.

Note: Solving this differential equation analytically gives a different answer, $\displaystyle 9.2997$. As the step size gets larger, Euler's method gives a more accurate answer. 

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,  

$\displaystyle f(x,y)=\frac{dy}{dx}=\frac{y}{x}$ 

$\displaystyle h=0.25$

Starting at the initial point $\displaystyle (x_0,y_0)=(1,1)$

$\displaystyle x_{1}=x_0+h=1+0.25=1.25$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=1+0.25*(1/1)=1.25$

$\displaystyle (x_1,y_1)=(1.25,1.25)$

$\displaystyle x_{2}=x_1+h=1.25+0.25=1.5$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=1.25+0.25*(1.25/1.25)=1.5$

$\displaystyle (x_2,y_2)=(1.5,1.5)$

We continue using Euler's method until $\displaystyle x=2$.  The results of Euler's method are in the table below.

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,

$\displaystyle f(x,y)=\frac{dy}{dx}=\frac{y}{x}$ 

$\displaystyle h=0.1$

Starting at the initial point $\displaystyle (x_0,y_0)=(1,1)$

$\displaystyle x_{1}=x_0+h=1+0.1=1.1$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=1+0.1*(1/1)=1.1$

$\displaystyle (x_1,y_1)=(1.1,1.1)$

$\displaystyle x_{2}=x_1+h=1.1+0.1=1.2$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=1.1+0.1*(1.1/1.1)=1.2$

$\displaystyle (x_2,y_2)=(1.2,1.2)$

We continue using Euler's method until $\displaystyle x=2$.  The results of Euler's method are in the table below.

Note: Due to the simplicity of the differential equation, Euler's method finds the exact solution, even with a large step size,  Using a smaller step size is unnecessary and more time consuming.

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,

$\displaystyle f(x,y)=\frac{dy}{dx}=cos(x)$ 

$\displaystyle h= \pi/10$

Starting at the initial point $\displaystyle (x_0,y_0)=(0,0)$

$\displaystyle x_{1}=x_0+h=0+\pi/10=\pi/10$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0+(\pi/10)*cos(0)=\pi/10$

$\displaystyle (x_1,y_1)=(\pi/10,\pi/10)$

$\displaystyle x_{2}=x_1+h=\pi/10+\pi/10=\pi/5$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=\pi/10+(\pi/10)*cos(\pi/10)=\pi/5$

$\displaystyle (x_2,y_2)=(\pi/5,\pi/5)$

We continue using Euler's method until $\displaystyle x=\pi/2$.  The results of Euler's method are in the table below.

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,

$\displaystyle f(x,y)=\frac{dy}{dx}=cos(x)$ 

$\displaystyle h= \pi/20$

Starting at the initial point $\displaystyle (x_0,y_0)=(0,0)$

$\displaystyle x_{1}=x_0+h=0+\pi/20=\pi/20$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0+(\pi/20)*cos(0)=\pi/20$

$\displaystyle (x_1,y_1)=(\pi/20,\pi/20)$

$\displaystyle x_{2}=x_1+h=\pi/20+\pi/20=\pi/10$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=\pi/20+(\pi/20)*cos(\pi/20)=\pi/10$

$\displaystyle (x_2,y_2)=(\pi/10,\pi/10)$

 We continue using Euler's method until $\displaystyle x=\pi/2$.  The results of Euler's method are in the table below.

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,

$\displaystyle f(x,y)=\frac{dy}{dx}=y^2e^{x}$ 

$\displaystyle h= 1$

Starting at the initial point $\displaystyle (x_0,y_0)=(0,0.01)$

$\displaystyle x_{1}=x_0+h=0+1=1$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0.01+1*(0.01)^2*e^{0}=0.0101$

$\displaystyle (x_1,y_1)=(1,0.0101)$

$\displaystyle x_{2}=x_1+h=1+1=2$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=0.0101+1*(0.0101)*e^{1}=0.01038$

$\displaystyle (x_2,y_2)=(2,0.01038)$

We continue using Euler's method until $\displaystyle x=6$.  The results of Euler's method are in the table below.

Euler's method uses iterative equations to find a numerical solution to a differential equation.  The following equations

$\displaystyle x_{n+1}=x_n+h$

$\displaystyle y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value.  $\displaystyle f(x_n,y_n)$ is the solution to the differential equation.

In this problem,

$\displaystyle f(x,y)=\frac{dy}{dx}=y^2e^{x}$ 

$\displaystyle h= 0.5$

Starting at the initial point $\displaystyle (x_0,y_0)=(0,0.01)$

$\displaystyle x_{1}=x_0+h=0+0.5=0.5$

$\displaystyle y_{1}=y_0+h*f(x_0,y_0)=0.01+0.5*(0.01)^2*e^{0}=0.01005$

$\displaystyle (x_1,y_1)=(0.5,0.01005)$

$\displaystyle x_{2}=x_1+h=0.5+0.5=1$

$\displaystyle y_{2}=y_1+h*f(x_1,y_1)=0.01005+0.5*(0.01005)*e^{1}=0.01013$

$\displaystyle (x_2,y_2)=(1,0.01013)$

We continue using Euler's method until $\displaystyle x=6$.  The results of Euler's method are in the table below.