# Differential equations, euler's method and bisection method

• meddi83
In summary, the conversation discusses a problem involving a second-order ODE and the use of Euler's method and the bisection method to find an accurate value of y(Iab). The problem is broken down into two first-order ODEs and the steps for solving using Euler's method are outlined. The conversation also addresses a typo in the problem and clarifies the use of Euler's method for finding y values at different points.
meddi83

## Homework Statement

Hi guys, I have the following problem and I don't know how to start.

I am given that W = 0.5, X(0) = 0, Iab = 20m and ha = 5m $$\frac{dy}{dx}=0$$

and

$$\frac{dy^2}{d^2x}=\frac{W}{T}\sqrt{1 + (\frac{dy}{dx})^2}$$

I am told to convert the 2nd order ODE to two 1st order ODEs at first. Then I am asked to choose an arbitrary value of T and integrate the system of equations using Euler's method from 0-Iab, where Iab=20, with 3 different step sizes. Then, I have to select the step size that gives an accurate value of y(Iab) (i.e. approximate relative error less than 0.1% if comparing with ha = 5).. Then I am asked to chose two values of T such that one gives y(Iab) smaller than ha and the other one gives y(Iab) bigger than ha and use those two values with the bisection method

## Homework Equations

Euler's method: $$yi+1 = yi + f(xi,yi)*h$$ where $$f(x,y) = \frac{dy}{dx}$$

## The Attempt at a Solution

At first I broke it to two 1st order ODES:

$$\frac{dy}{dx}=g$$ and $$\frac{dg}{dx}=\frac{W}{T}\sqrt{1 + g^2}$$

Then, for Euler's method: $$yi+1 = yi + gi*h$$ and $$gi+1 = gi + (\frac{W}{T}\sqrt{1+gi^2})*h$$

My problem is, I don't know where to start in order to find y(Iab) (because the problem is asking for an accurate value of y(Iab) )... and what step size (h) to use. .. I also don't know which equation shall be used with the bisection method..

any help would be really appreciated. thank you

I would suggest you go back and reread the problem. You say that X(0)= 0 but your differential equation has y as a function of x. What does X(0) mean?

Sorry, typo mistake. It says that the slope, at x=0, is dy/dx=0

and from the Picture I have, the graph, when x=0 -> y=0.

So I suppose for euler's method, suppose I choose a step size of ie, h=5, then it would be

$$y(5) = y(0) + g(0)*h$$ and $$g(5) = g(0) + (\frac{W}{T}\sqrt{1+g(0)^2})*h$$

Is that correct ? And since I have to integrate from 0-20, I suppose I'll have to do y(10), y(15) and y(20) aswell, right ? (Same for g(10),g(15),g(20) ) ?

## 1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model various physical phenomena and can be solved to determine the behavior of the function over time or space.

## 2. What is Euler's method?

Euler's method is a numerical method for solving differential equations. It involves dividing the interval of interest into smaller subintervals and using the slope of the function at a given point to approximate the value of the function at the next point.

## 3. How does Euler's method work?

Euler's method works by using the derivative of a function to approximate the value of the function at a given point. This is done by taking small steps along the function, using the slope of the function at each point to determine the next point, and repeating this process until the desired accuracy is achieved.

## 4. What is the bisection method?

The bisection method is a numerical method for finding the root of a function. It involves repeatedly dividing the interval of interest in half and checking which half contains the root. This process is repeated until the root is found with the desired accuracy.

## 5. How does the bisection method work?

The bisection method works by dividing the interval of interest in half and checking which half contains the root. This process is repeated until the root is found with the desired accuracy. It is based on the intermediate value theorem, which states that if a continuous function has different signs at the endpoints of an interval, then it must have a root within that interval.

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