Euler's method to estimate y(1) for y' = 1 - xy, y(0) = 0

[paulxzt]

Use Euler's method with step size .2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = 1-xy, y(0)=0.

Can someone help me with this.. i looked over the example in the book and still can't seem to figure this one out. I tried another problem above this problem and checked the back to see if i got it but i can't seem to get the same answer.

 

[tkhunny]

Was the answer elsewhere unsatisfactory? All you need is the Point-Slope form of a line. It is not more difficult than that.

The point is (0,0)
The slope at (0,0) is y' = 1-(0)(0) = 1
The equation of the linear approximating line at (0,0) is (y-0) = 1*(x-0) or y = x
Using the line y = x, move x out to x = 0.2.
The new y-value is y = 0.2, since y = x
The next point is (0.2,0.2)

Now start over.

The point is (0.2,0.2)
The slope at (0.2,0.2) is y' = 1-(0.2)(0.2) = 1 - 0.04 = 0.96
The equation of the linear approximating line at (0.2,0.2) is (y-0.2) = 0.96*(x-0.2) or y = 0.96(x-0.2)+0.2
Using the line y = 0.96*(x-0.2)+0.2, move x out to x = 0.4.
The new y-value is y = 0.392, since y = 0.96*(x-0.2)+0.2
The next point is (0.4,0.392)

Now start over.
The point is (0.4,0.392)
...

Notice how the x value moves up by 0.2 each time. This is what "stepsize = 0.2" means.

You can follow the rote and laborious procedure above, or you can switch on your algebra brain and simplify the arithmetic by creating an interative function as shown in your other answer to this question.