Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem below. Give your answer correct to 4 decimal places.

y' = 1 - xy
y(0) = 0

y(1) = ____ ?

... help, this is what i've done but got the wrong answer..

h = 0.2
x0 = 0
y0 = 0
F(x,y) = 1-xy

y1 = y0 + hF(x0,y0)
= 0 + 0.2(1-(0)(0)) = 0.2

y2 = y1 + hF(x1,y1)
= 0.2 + 0.2(1-(0.2)(0.2)) = 0.392

y3 = y2 + hF(x2,y2)
= 0.392 + 0.2(1-(0.4)(0.392)) = 0.5606

y4 = y3 + hF(x3,y3)
= 0.5606 + 0.2(1-(0.6)(0.5606)) = 0.6934

y5 = y4 + hF(x4,y4)
= 0.6934 + 0.2(1-(0.8)(0.6934)) = 0.7824

y6 = y5 + hF(x5,y5)
= 0.7824 + 0.2(1-(1.0)(0.7824)) = 0.8259

...my answer was 0.8259, which was wrong. please show me what i'm doing wrong? thanks!

y5 is the estimate for y(1), not y6:

y5 = 1528174/1953125 = 0.782425088

The exact answer involves the error function of imaginary argument and is to 30 significant figures:

0.724778459007076331818227967606

You haven't done anything wrong in your work, you have to remember that as you are solving for your y's (y1, y2, y3... etc.) you are also solving for the corresponding x-values (x1, x2, x3... etc.) therefore when your x-value (x5) =1 your corresponding y-value (y5)= .7824

It seems like you made a mistake in calculating y5. Let's go through the steps again to find the correct answer.

Given:
h = 0.2
x0 = 0
y0 = 0
F(x,y) = 1 - xy

Using Euler's method, we have:
y1 = y0 + h * F(x0, y0)
= 0 + 0.2 * (1 - (0)(0))
= 0.2

y2 = y1 + h * F(x1, y1)
= 0.2 + 0.2 * (1 - (0.2)(0.2))
= 0.2 + 0.2 * (1 - 0.04)
= 0.2 + 0.2 * 0.96
= 0.2 + 0.192
= 0.392

y3 = y2 + h * F(x2, y2)
= 0.392 + 0.2 * (1 - (0.4)(0.392))
= 0.392 + 0.2 * (1 - 0.156544)
= 0.392 + 0.2 * 0.843456
= 0.392 + 0.1686912
= 0.5606912

y4 = y3 + h * F(x3, y3)
= 0.5606912 + 0.2 * (1 - (0.6)(0.5606912))
= 0.5606912 + 0.2 * (1 - 0.33641472)
= 0.5606912 + 0.2 * 0.66358528
= 0.5606912 + 0.132717056
= 0.693408256

y5 = y4 + h * F(x4, y4)
= 0.693408256 + 0.2 * (1 - (0.8)(0.693408256))
= 0.693408256 + 0.2 * (1 - 0.5547266048)
= 0.693408256 + 0.2 * 0.4452733952
= 0.693408256 + 0.08905467904
= 0.78246293504

Therefore, the correct estimate for y(1) using Euler's method with step size 0.2 is approximately 0.7825 (rounded to 4 decimal places).

In Euler's method, the formula for approximating the next value, y(x_i+1), given the current value, y(x_i), is:

y(x_i+1) ≈ y(x_i) + h * F(x_i, y(x_i))

Where h is the step size and F(x,y) is the derivative function.

From your calculation, it seems that you have correctly set up the initial conditions:

h = 0.2
x0 = 0
y0 = 0

The derivative function is given as F(x,y) = 1 - x * y.

Now, let's go through the calculation step by step:

Step 1:
x1 = x0 + h = 0 + 0.2 = 0.2
y1 ≈ y0 + h * F(x0, y0) = 0 + 0.2(1 - (0)(0)) = 0.2

Step 2:
x2 = x1 + h = 0.2 + 0.2 = 0.4
y2 ≈ y1 + h * F(x1, y1) = 0.2 + 0.2(1 - (0.2 * 0.2)) = 0.2 + 0.2(1 - 0.04) = 0.2 + 0.2(0.96) = 0.392

Step 3:
x3 = x2 + h = 0.4 + 0.2 = 0.6
y3 ≈ y2 + h * F(x2, y2) = 0.392 + 0.2(1 - (0.4 * 0.392)) = 0.392 + 0.2(1 - 0.156) = 0.392 + 0.2(0.844) = 0.392 + 0.1688 = 0.5608

Step 4:
x4 = x3 + h = 0.6 + 0.2 = 0.8
y4 ≈ y3 + h * F(x3, y3) = 0.5608 + 0.2(1 - (0.6 * 0.5608)) = 0.5608 + 0.2(1 - 0.3365) = 0.5608 + 0.2(0.6635) = 0.5608 + 0.1327 = 0.6935

Step 5:
x5 = x4 + h = 0.8 + 0.2 = 1.0
y5 ≈ y4 + h * F(x4, y4) = 0.6935 + 0.2(1 - (0.8 * 0.6935)) = 0.6935 + 0.2(1 - 0.5548) = 0.6935 + 0.2(0.4452) = 0.6935 + 0.089 = 0.7825

Therefore, using Euler's method with a step size of 0.2, the estimate for y(1) is approximately 0.7825.