Euler s Method Continuous Functions and Field Theory

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Suppose we have the following differential equation with the initial condition:

$\frac{\partial p }{\partial x} = 0.5x(1-x), \ p(0) = 2$

Use Euler's method to approximate p(2) , using a step size of .

Correct answer:

Explanation:

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:

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

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

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

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

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

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

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

Approximate $\small e^{\frac{1}{2}}$ by using Euler's method on the differential equation

$\small y'=y$

with initial condition $\small y(0)=1$ (which has the solution $\small y(t)=e^t$ ) and time step $\small \small h=\frac{1}{4}$ .

Correct answer:

$\small 1.5625$

Explanation:

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

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

where each $\small t_i$ is

$\small t_0=0$

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

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

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

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

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

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

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

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=3x+4y$ at x=1 with the initial condition y(0)=0 and step size h=0.25 .

Correct answer:

(1,2.0625)

Explanation:

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

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

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

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

In this problem,

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

h=0.25

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

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

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

(x_1,y_1)=(0.25,0)

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

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

(x_2,y_2)=(0.5,0.1875)

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

Screen shot 2015 11 19 at 3.48.44 pm

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

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=3x+4y$ at x=1 with the initial condition y(0)=0 and step size h=0.1 .

Correct answer:

(1,4.4860)

Explanation:

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

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

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

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

In this problem,

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

h=0.1

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

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

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

(x_1,y_1)=(0.1,0)

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

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

(x_2,y_2)=(0.2,0.03)

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

Problem 2

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

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=\frac{y}{x}$ at x=2 with the initial condition y(1)=1 and step size h=0.25 .

Correct answer:

(2,2)

Explanation:

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

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

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

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

In this problem,

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

h=0.25

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

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

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

(x_1,y_1)=(1.25,1.25)

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

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

(x_2,y_2)=(1.5,1.5)

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

Problem 3

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=\frac{y}{x}$ at x=2 with the initial condition y(1)=1 and step size h=0.1 .

Correct answer:

(2,2)

Explanation:

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

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

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

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

In this problem,

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

h=0.1

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

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

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

(x_1,y_1)=(1.1,1.1)

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

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

(x_2,y_2)=(1.2,1.2)

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

Problem 4

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.

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=cos(x)$ at $x=\pi/2$ with the initial condition y(0)=0 and step size $h= \pi/10$ .

Correct answer:

$(\pi/2,1.14)$

Explanation:

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

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

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

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

In this problem,

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

$h= \pi/10$

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

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

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

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

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

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

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

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

Problem 7

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=cos(x)$ at $x=\pi/2$ with the initial condition y(0)=0 and step size $h= \pi/20$ .

Correct answer:

$(\pi/2,1.07)$

Explanation:

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

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

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

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

In this problem,

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

$h= \pi/20$

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

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

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

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

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

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

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

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

Problem 8

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=y^2e^{x}$ at x=6 with the initial condition y(0)=0.01 and step size h= 1 .

Correct answer:

(6,0.1089)

Explanation:

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

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

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

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

In this problem,

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

h= 1

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

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

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

(x_1,y_1)=(1,0.0101)

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

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

(x_2,y_2)=(2,0.01038)

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

Problem 11

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=y^2e^{x}$ at x=6 with the initial condition y(0)=0.01 and step size h= 0.5 .

Correct answer:

(6,5.113)

Explanation:

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

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

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

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

In this problem,

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

h= 0.5

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

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

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

(x_1,y_1)=(0.5,0.01005)

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

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

(x_2,y_2)=(1,0.01013)

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

Problem 12

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Euler s Method Continuous Functions and Field Theory

All Calculus 2 Resources

All Calculus 2 Resources

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