00:01
We are given an initial value problem, and we are asked to use euler's method with a given step size to approximate some points on the solution.
00:18
The initial value problem is the differential equation y prime equals y minus 2x, with initial value y of 1 equals 0, and the step size we are asked to use is h equals 0 .5.
00:35
Since our initial value was y of 1 equals 0, we have that x0 is equal to 1, and y0 is equal to 0.
00:49
Therefore, we have that x1 is equal to x0 plus 0 or 1 .5, and y1 is equal to y0, which is 0, plus our step size, 0, times our function, y, 05 .5, times our function, y, minus 2 .5 .5 .5.
01:14
X evaluated at point x0 y0 or point one zero this is simply equal to negative 2 times 1 which is equal to 0 .5 times negative 2 is negative 1 this is negative 1 again because our step size is 0 .5 we have the x2 is equal to 2 .0 and therefore y2 is equal to y2 is equal to y2 is equal to y which is negative 1 plus our step size 0 .5 times our function y minus 2x evaluated at x1y1 or at 1 .5 negative 1.
02:11
This is equal to negative 1 minus 2 times 1 .5 which is equal to negative 1 plus 0 .5 times negative 1 minus 3 or negative 4, which is equal to negative 1 plus negative 2 or negative 3.
02:43
For our third step, we have that x3 is equal to 2 plus 0 .5, or 2 .5, and that y of 3 is equal to y2, which is negative 3, plus our step size 0 .5 times our function y minus 2x, evaluated at the point 2, negative 3...