Let y ′ ( t ) = y 2 y^{\prime}\left(t\right)=\frac{y}{2} with y ( 0 ) = 2 y\left(0\right)=2 . Compute the first three approximations given by Euler’s Method with a step size of h = 0.2 h=0.2 .

2.2 , 2.42 , 2.662 2.2,2.42,2.662

13. Intro to Differential Equations

Euler's Method

13. Intro to Differential Equations

Euler's Method: Videos & Practice Problems

Topic summary

Euler's method enhances function value approximation by utilizing multiple shorter tangent lines instead of a single tangent line. Starting with an initial condition, the method calculates new x and y values through iterative steps, incorporating the derivative of the function. For example, given the differential equation y' = 2x and an initial point (1, 2), using a step size of h = 0.5, the approximated function value at x = 2 is 4.5, demonstrating improved accuracy over linear approximation. Smaller step sizes can further refine this approximation.

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Euler's Method

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Euler's Method Video Summary

Euler's method is a numerical technique used to approximate the values of a function based on its derivative, providing a more accurate representation than linear approximation alone. This method involves using multiple shorter tangent lines, which allows the approximation to closely follow the curve of the function.

To illustrate Euler's method, consider the differential equation $ y' = 2x $ with the initial condition $ (x_0, y_0) = (1, 2) $. Our goal is to approximate $ f(2) $ using two steps of size $ h = 0.5 $.

Initially, we can use the linear approximation to estimate $ f(2) $. By applying the linearization formula, we find that $ f(2) $ is approximately 4. However, this estimate is not very close to the actual function curve, highlighting the need for a more refined approach.

In Euler's method, we start with the initial condition and organize our calculations in a table. The first step involves calculating the new $ x $ and $ y $ values:

1. For $ x_1 $: \[ x_1 = x_0 + h = 1 + 0.5 = 1.5 \]

2. For $ y_1 $: \[ y_1 = y_0 + h \cdot f'(x_0) = 2 + 0.5 \cdot (2 \cdot 1) = 2 + 1 = 3 \]

Now we plot the point $ (1.5, 3) $ on the graph and connect it to the initial point. Next, we take another step to find $ x_2 $ and $ y_2 $:

1. For $ x_2 $: \[ x_2 = x_1 + h = 1.5 + 0.5 = 2 \]

2. For $ y_2 $: \[ y_2 = y_1 + h \cdot f'(x_1) = 3 + 0.5 \cdot (2 \cdot 1.5) = 3 + 1.5 = 4.5 \]

We plot the point $ (2, 4.5) $ and connect it to the previous point. Thus, we have approximated $ f(2) $ as 4.5, which is significantly closer to the actual function curve than our initial linear approximation.

It is important to note that the accuracy of Euler's method can be improved by using smaller step sizes, such as $ h = 0.25 $ or $ h = 0.1 $. This results in a more precise approximation of the function values. As you continue to practice Euler's method, you will gain a deeper understanding of its application and effectiveness in solving differential equations.

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Euler's Method Example 1

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Euler's Method Example 1 Video Summary

To estimate the solution of the differential equation $ y' = 2x - y $ with the initial condition $ y(0) = -1 $ over the interval from 0 to 1 using Euler's method, we first determine the step size $ h $. Given that we are using 5 steps, we calculate $ h $ as follows:

\[h = \frac{1 - 0}{5} = 0.2\]

Next, we set up a table to track the values of $ x_n $ and $ y_n $. Starting with the initial values:

\[x_0 = 0, \quad y_0 = -1\]

We can compute the subsequent $ x $ values by adding the step size $ h $ to the previous $ x $ value:

\[\begin{align*}x_1 & = x_0 + h = 0 + 0.2 = 0.2 \\x_2 & = x_1 + h = 0.2 + 0.2 = 0.4 \\x_3 & = x_2 + h = 0.4 + 0.2 = 0.6 \\x_4 & = x_3 + h = 0.6 + 0.2 = 0.8 \\x_5 & = x_4 + h = 0.8 + 0.2 = 1.0\end{align*}\]

Now, we calculate the $ y $ values using the formula:

\[y_n = y_{n-1} + h \cdot f(x_{n-1}, y_{n-1})\]

where $ f(x, y) = 2x - y $. We proceed to find each $ y_n $:

For $ n = 1 $:

\[y_1 = y_0 + h \cdot f(x_0, y_0) = -1 + 0.2 \cdot (2 \cdot 0 - (-1)) = -1 + 0.2 \cdot 1 = -0.8\]

For $ n = 2 $:

\[y_2 = y_1 + h \cdot f(x_1, y_1) = -0.8 + 0.2 \cdot (2 \cdot 0.2 - (-0.8)) = -0.8 + 0.2 \cdot (0.4 + 0.8) = -0.8 + 0.2 \cdot 1.2 = -0.56\]

For $ n = 3 $:

\[y_3 = y_2 + h \cdot f(x_2, y_2) = -0.56 + 0.2 \cdot (2 \cdot 0.4 - (-0.56)) = -0.56 + 0.2 \cdot (0.8 + 0.56) = -0.56 + 0.2 \cdot 1.36 = -0.288\]

For $ n = 4 $:

\[y_4 = y_3 + h \cdot f(x_3, y_3) = -0.288 + 0.2 \cdot (2 \cdot 0.6 - (-0.288)) = -0.288 + 0.2 \cdot (1.2 + 0.288) = -0.288 + 0.2 \cdot 1.488 = 0.0096\]

For $ n = 5 $:

\[y_5 = y_4 + h \cdot f(x_4, y_4) = 0.0096 + 0.2 \cdot (2 \cdot 0.8 - 0.0096) = 0.0096 + 0.2 \cdot (1.6 - 0.0096) = 0.0096 + 0.2 \cdot 1.5904 = 0.32768\]

Thus, the approximation of $ y(1) $ using Euler's method is approximately $ 0.32768 $.

To compare this with the exact solution, we use the formula:

\[y = 2x + e^{-x} - 2\]

Substituting $ x = 1 $:

\[y(1) = 2 \cdot 1 + e^{-1} - 2 = 2 + \frac{1}{e} - 2 = \frac{1}{e} \approx 0.367879\]

The difference between the exact and approximate values is:

\[\text{Difference} = |0.367879 - 0.32768| \approx 0.040199\]

This analysis shows that the approximation using Euler's method is fairly close to the exact solution, with a difference of approximately $ 0.0402 $.

Problem

Use Euler’s method with a step size of $h equals 0.5$ to estimate the value of $y of 2$ , where $y$ is the solution of the initial value problem $y^{'} = 2 x, y (0) = 1$

$1.5$

$1$

$2.5$

$4$

Problem

Let $y^{'} (t) = \frac{y}{2}$ with $y of 0 equals 2$ . Compute the first three approximations given by Euler’s Method with a step size of $h equals 0.2$ .

$2 comma 2.2 comma 2.42$

$2.2 comma 2.42 comma 2.662$

$0 comma 0.2 comma 0.4$

$0.2 comma 0.4 comma 0.6$

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What is Euler's Method and how is it used in calculus?

Euler's Method is a numerical technique used in calculus to approximate solutions to differential equations. Instead of relying on a single tangent line for approximation, it uses multiple shorter tangent lines to follow the curve of the function more closely. The method starts with an initial condition, such as $x_{0}$ and $y_{0}$ , and iteratively calculates new points using the formulas:

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

$y_{n} = y_{n-1} + h * f'(x_{n-1}, y_{n-1})$

where $h$ is the step size and $f'$ is the derivative. By repeating this process, Euler's Method provides an approximation of the function's value at a desired point.

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How does step size affect the accuracy of Euler's Method?

The step size $h$ in Euler's Method significantly impacts the accuracy of the approximation. Smaller step sizes result in more steps and shorter tangent lines, which better follow the curve of the actual function, leading to a more accurate approximation. However, smaller step sizes also increase the computational effort. Conversely, larger step sizes reduce the number of steps but may deviate more from the true function, resulting in less accurate approximations. For example, using $h = 0.5$ might yield an approximation of $f(2)$ as 4.5, while a smaller $h$ like 0.25 could provide an even closer value to the actual function.

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What are the steps to apply Euler's Method to a differential equation?

To apply Euler's Method to a differential equation, follow these steps:

Start with the initial condition, $(x_{0}, y_{0})$ .
Determine the step size $h$ .
Use the formulas to calculate the next values:

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

$y_{n} = y_{n-1} + h * f'(x_{n-1}, y_{n-1})$

Repeat the process for the desired number of steps or until the target $x$ value is reached.
Plot the points if needed to visualize the approximation.

This iterative process provides an approximation of the solution to the differential equation.

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What is the difference between linear approximation and Euler's Method?

Linear approximation uses a single tangent line at a specific point to estimate the value of a function near that point. It is based on the formula:

$l(x) = f(a) + f'(a)(x - a)$

where $a$ is the point of tangency. However, this method becomes less accurate as the distance from $a$ increases.

Euler's Method, on the other hand, uses multiple shorter tangent lines to approximate the function over a range of values. It iteratively calculates new points using the formulas:

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

$y_{n} = y_{n-1} + h * f'(x_{n-1}, y_{n-1})$

This approach provides a more accurate approximation, especially over larger intervals, as it accounts for the curvature of the function.

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Can Euler's Method be used for any differential equation?

Euler's Method can be applied to many first-order differential equations, but it has limitations. It works best for equations where the derivative $f'(x, y)$ is well-behaved (continuous and not excessively steep) over the interval of interest. For stiff equations or those with rapidly changing solutions, Euler's Method may require very small step sizes to maintain accuracy, which can make it computationally expensive. Additionally, it is less accurate compared to more advanced numerical methods like the Runge-Kutta methods. Despite these limitations, Euler's Method is a valuable tool for understanding the basics of numerical approximation and serves as a foundation for more complex techniques.

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Euler's Method: Videos & Practice Problems

Euler's Method

Euler's Method Video Summary

Euler's Method Example 1

Euler's Method Example 1 Video Summary

Use Euler’s method with a step size of h=0.5h=0.5 to estimate the value of y(2)y\left(2\right), where yy is the solution of the initial value problem y′=2x,y(0)=1y^{\prime}=2x,y\left(0\right)=1

Let y′(t)=y2y^{\prime}\left(t\right)=\frac{y}{2} with y(0)=2y\left(0\right)=2. Compute the first three approximations given by Euler’s Method with a step size of h=0.2h=0.2.

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Here’s what students ask on this topic:

What is Euler's Method and how is it used in calculus?

How does step size affect the accuracy of Euler's Method?

What are the steps to apply Euler's Method to a differential equation?

What is the difference between linear approximation and Euler's Method?

Can Euler's Method be used for any differential equation?

Your Calculus tutors

Use Euler’s method with a step size of $h equals 0.5$ to estimate the value of $y of 2$ , where $y$ is the solution of the initial value problem $y^{'} = 2 x, y (0) = 1$

Let $y^{'} (t) = \frac{y}{2}$ with $y of 0 equals 2$ . Compute the first three approximations given by Euler’s Method with a step size of $h equals 0.2$ .