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.

2.2 , 2.42 , 2.662 2.2,2.42,2.662

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 .

we're told to let y prime of t equal y over two with an initial condition of y of zero being equal to two. We're asked to compute the first three approximations given by Euler's method with a step size of h equals 0.2. And because we're using Euler's method here, the first thing I'm gonna do is just set up a table of values with n, x sub n, and y seven. Now given our initial condition that y of zero is equal to two, I can go ahead and fill that in here for n being a zero. So that zero in parentheses represents x sub zero and that two is y sub zero. So I can fill that in here. And we're asked to compute the first three approximations given by Euler's method. So I can go ahead and fill in one, two, and three for n here. Now one thing to note is that we are actually working with a function of t rather than a function of x, but it will work out the same. So I am gonna go ahead and replace this x with a t just to be a bit more specific to this problem. So I have t sub n, and this initial condition is based on t sub zero. So now I can easily find what those t values are based on my first equation here. We just wanna take our previous t value value and add our step size. Our step size was given to us as 0.2. So adding zero plus 0.2 gives me 0.2. Then adding another point two gives me point four and another gives me point six. Just adding that step size each time. Now we want to find y sub n starting with y one. Based on our equation here, we're going to take our previous y value, which here is two, and we're going to add that together with our step size 0.2 multiplied by the derivative of our function for our previous x and y values. So here, our derivative y prime is equal to y over two. So I really only need to worry about about that previous y value, which remember was two. So this is two over two, just plugging in that previous y value to our derivative. Two over two is one, so this is just two plus point two, which will give us 2.2. So filling that in my table here, we can go ahead and find a y two again basing that on our previous y value. So we're taking 2.2. We're adding that together with 0.2 times 2.2 over two. That will give us here 2.42. And then finally, y three. That's going to be our previous Y value, 2.42 plus 0.2 times 2.42 over two. Again, plugging that previous Y value into our derivative and this will give us here 2.662 for that last value. So this gives us here our first three approximations for Euler's method based on all the information given to us in the problem. Feel free to let me know if you have any questions and I'll see you in the next one.

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

Calculus

13. Intro to Differential Equations

Euler's Method

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Multiple Choice

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 $h equals 0.2$ .

$2 comma 2.2 comma 2.42$

$2.2 comma 2.42 comma 2.662$

$0 comma 0.2 comma 0.4$

$0.2 comma 0.4 comma 0.6$

Verified step by step guidance

Step 1: Understand the problem. Euler's Method is a numerical technique to approximate solutions to differential equations. The formula for Euler's Method is y_{n+1} = y_n + h * f(t_n, y_n), where h is the step size, and f(t, y) is the derivative y′(t). Here, y′(t) = y/2, the initial condition is y(0) = 2, and the step size is h = 0.2.

Step 2: Start with the initial condition. At t_0 = 0, y_0 = 2. Use the formula y_{n+1} = y_n + h * f(t_n, y_n) to compute the next value. For the first step, t_1 = t_0 + h = 0 + 0.2 = 0.2, and y_1 = y_0 + h * (y_0 / 2) = 2 + 0.2 * (2 / 2).

Step 3: Compute the second approximation. Using the updated values, t_2 = t_1 + h = 0.2 + 0.2 = 0.4. For y_2, substitute y_1 into the formula: y_2 = y_1 + h * (y_1 / 2).

Step 4: Compute the third approximation. Update t_3 = t_2 + h = 0.4 + 0.2 = 0.6. For y_3, substitute y_2 into the formula: y_3 = y_2 + h * (y_2 / 2).

Step 5: Summarize the results. The first three approximations for y(t) using Euler's Method with step size h = 0.2 are y_1, y_2, and y_3, corresponding to t = 0.2, 0.4, and 0.6, respectively. These values are obtained by iteratively applying the formula.

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