00:01
Okay, so we're given this differential equation here, y prime is equal to y minus 2x, with the initial condition, y1 is equal to 0.
00:08
So we need to use euler's method with h is equal to 0 .5 and approximate y1, y2, y3, and y4.
00:18
So we're going to use four iterations of this, of euler's method.
00:25
So euler's method says that y of i is equal to y of i.
00:30
Minus 1 plus h times f of x i minus 1 times y i minus 1.
00:37
Here f of x y is equal to y minus 2 x right then also we have that x i should just be x i minus 1 plus h.
00:51
So let's have x i y i i here we have 0 1 2 3 and then 4 so xi is going to be 1, and then yi is 0.
01:06
So let's go ahead and fill in all of the xi, so that's going to be 1 .5, 2 .0, 2 .5, and then 3 .0 here.
01:16
Now, yi, so we'll do y1.
01:19
Y1 is going to be y0 plus h times, and then we have y0 minus 2 times x of 0.
01:29
Plugging all of that in, y of 0 is 0 plus 0 .5 times 0 minus 2 times 1.
01:40
So here, now we get, okay, so we get negative 1 here.
01:46
So just 0 .5 times negative 2...