00:01
So, we have given here the derivative which is dy by dx that is equal to, we need to write here, 1 plus x upon y.
00:09
And the initial condition we have given at 1, y gives a value 1.
00:14
So now by euler method, the euler method we need to apply here the interval x is less than equal to 4 and greater than equal to 1.
00:23
So which is the euler method is yn plus 1 is equal to yn plus h into f of xn, yn.
00:38
So first, from the first step we need to find here y1 which is equal to y0 plus h and that is f of x0, y0.
00:51
So from here we will get that is 1 plus h value is we have here 0 .5 and that is f of 1, 1.
01:04
So which is we have here that is 1 plus 0 .5 into 1 plus 1.
01:12
So from here we will get that is our answer become.
01:16
So from here our answer become y1 is equal to 2.
01:23
So then we can write x1 is equal to x0 plus h.
01:30
So from here x1 is equal to we have to write that is 1 plus 0 .5 that is equal to 1 .5.
01:40
Now similarly according to a second step we need to find here that is y2 is equal to y1 plus h of f function that is x1, y1...