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

we're asked to use Euler's method with a step size of 0.5 to estimate the value of y of two, so our function value when x is two, where y is a solution to the initial value problem, a y prime equals two x, where y of zero is equal to one. Now because we're asked to use Euler's method here, the first thing I'm gonna do is set up a table where I have n, x seven, and y seven. Now I know that when n is zero, that's my initial condition, x sub zero and y sub zero, which were given to me in the the problem. This zero here is x zero and then this one is y zero. So I can go ahead and plug that in right here or rather write it in my table. And we're given our step size of point five here and we're asked to estimate y of two. So remembering that x sub n, we find this by just adding our step size to our previous x value. I'm going to start with zero, then adding point five. I have a 0.5, then I have a one, then a 1.5, then two, because that's what we're trying to estimate here. That's how far I wanna go. So I have one, two, three, four steps. I'm gonna go ahead and draw these lines in just to finish out my table. And now we want to determine what those y values are. So starting here with y one, using our equation here, we're going to take our previous y value, which here was one. So taking one, adding it together with our step size h of 0.5 and multiplying that by our derivative here, which is two x based on my previous x and y values. Because this only depends on x, I only need to worry about my previous x value, which is zero. So this is two times zero, two times zero zero. That whole term is just zero, so this ends up just being one. Now in order to find y two, we again take our previous Y value here. That's one. We add that together with our step size 0.5 times our derivative, which is two times our previous X value, which here was 0.5. So this is two times 0.5. Two times 0.5 is one. So this ends up being one plus 0.5, which is 1.5. So that's my next value here. Then for y three, that's going to be my previous y value, 1.5 plus my step size of 0.5 times two times my previous x value of one. Two times one is two. Two times point five is one. One point five plus one gives us here 2.5. Then we have one final Y value to find, which is going to be our previous Y value of 2.5 plus our step size of 0.5 times two times our previous X value of 1.5. Two times 1.5 is three. Three times 0.5 is 1.5. One point five plus 2.5 will give us four. So that gives us our estimate here for Y of two. So here we're estimating that y of two is going to be equal to four based on using Euler's method. And I can go ahead and write this out as y of two equals four. Feel free to let us know if you have any questions, and I'll see you in the next one.

13. Intro to Differential Equations

Euler's Method

Calculus

13. Intro to Differential Equations

Euler's Method

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Multiple Choice

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 $y^{'} = 2 x, y (0) = 1$

$1.5$

$1$

$2.5$

$4$

Verified step by step guidance

Step 1: Understand the problem. Euler's method is a numerical technique to approximate the solution of a differential equation. The given differential equation is y' = 2x with the initial condition y(0) = 1. We are tasked to estimate y(2) using a step size of h = 0.5.

Step 2: Recall the formula for Euler's method. The formula is y_{n+1} = y_n + h * f(x_n, y_n), where f(x, y) is the derivative given in the problem (in this case, f(x, y) = 2x). Start with the initial condition: x_0 = 0 and y_0 = 1.

Step 3: Perform the first iteration. Using the formula, calculate y_1: y_1 = y_0 + h * f(x_0, y_0). Substitute x_0 = 0, y_0 = 1, h = 0.5, and f(x_0, y_0) = 2(0).

Step 4: Perform the second iteration. Update x_1 = x_0 + h = 0 + 0.5 = 0.5. Then calculate y_2: y_2 = y_1 + h * f(x_1, y_1). Substitute x_1 = 0.5, y_1 (from the previous step), h = 0.5, and f(x_1, y_1) = 2(0.5).

Step 5: Perform the third iteration. Update x_2 = x_1 + h = 0.5 + 0.5 = 1. Then calculate y_3: y_3 = y_2 + h * f(x_2, y_2). Substitute x_2 = 1, y_2 (from the previous step), h = 0.5, and f(x_2, y_2) = 2(1). Continue this process until x = 2.