00:02
Hi, given, d .y by dx is equal to x cube by y, with y of 0 .3 is equal to 9.
00:11
Now here, fxy is this thing, x cube by y.
00:16
So this is my fxy, and this becomes my x not and this becomes my y -not.
00:23
And h is given to us as 0 .2.
00:26
So from euler's method, xn plus 1 is equal.
00:30
To x n plus h and y n plus 1 is equal to y n plus h f of x n y n so now for x 1 is equal to x0 plus h so this is equal to 0 .3 plus 0 .2 which is equal to 0 .5 and for y 1 this becomes y0 plus h of f of x not y not so y not is 9 plus h is 0 .0 plus h is 0 .0.
01:01
Into f of x not y not so putting x not x not is 0 .3 so this becomes 0 .3 cube divided by y not which is 9 so this simplification comes out to be equal to 9 .000 0 .06 6 so in similar ways we can find other values also so we have to fill the table n x n and then y n so for n 0 1 2 3 4 4, 5, xn is 0 .3, 0 .5, 0 .7, 0 .9, 1 .1 and 1 .3.
01:44
And yn becomes 9 in this case.
01:48
This is now yn values come out to be equal to 9 .006, 9 .0003, 9 .0003, 3 .3.
02:04
379 .01099, 9 .027 and 9 .0276...