00:01
Hello students in this question we have to apply the runga kuta method to the initial value problem which is dy over t x is equal to x plus y and we are given that at x equal to 0 y is 0 and choosing h equal to 0 .2 and computing 5 steps basically we can write it as y -dash is equal to x plus y which is f x comma y over here now we have to compute five steps so x h is equal to 0 .2 multiplied by 5 which is equal to 1 now according to the method we have x2 we have k2 is equal to h multiplied by f x0 plus h comma y0 plus k1 over here from here and we have know that k1 is h f x0 comma y not from here we have 0 .2 f of 0 comma 0 which is equal to 0 so from here we can find k2 is equal to h f this is here h f we will put the values if we have h s 0 .2 multiplied by we have f of x not so x not so x not is 0 plus h is 0 .2 so will be 0 .2 comma 0 .0 so this will come out to be 0 .0 4 we are putting the values of x at 0 .2 y is 0 so we are getting the values now here if we find y1 so y1 is equal to y0 plus k1 plus k2 over 2 so y0 is 0 plus when we have k1 which is 0 plus k2 is 0 .04 over 2.
02:13
So this will come out to be 0 .02.
02:16
So y1 is equal to 0 .02.
02:21
Now again, taking x1.
02:24
In place of x0.
02:25
X not, why not and repeat the process.
02:28
So we have k1 will be hf x1, x1, y1.
02:36
So from here we have 0 .2, mildly by way f of x1 is 0 .2 comma y1 is 0 .2 so this will come out to be 0 .04 4.
02:47
Similarly k2 we calculate as we calculated earlier k2 here we have k2 is equal to h f x1 plus h comma y1 plus k1 we have 0 .2 multiply f of this is x1 is 0 .2 plus h is 0 .2 plus h is 0 .2 so it will be 0 .04 this will be 0 .064 so when we find out with the help of function so it would be 0 .09 to 8 and similarly we find again y2 here y2 is equal to y1 plus k1 plus k2 over 2 now here this will come out to be 0 .0884 now now again we put x2 in place of x1.
03:54
So we have again k1 is equal to hf x2 comma y2.
04:03
Now this is 0 .2, f of x2 is 0 .4.
04:08
0 .0884.
04:12
So this comes out to be 0 .0977 over here.
04:17
And if we find k2 it will.
04:21
Be h f x2 plus h comma y2 plus k1 so this is equal to 0 .2 multiply f of x2 is 0 .4 plus h is 0 .2 so it will be 0 .6.
04:38
This will be 0 .186 1 so k2 will come out to be 0 .1572 so here we find y3 y3 will be y2 plus k1 plus k2 over 2 now y2 the value of y2 is 0 .084 so if we put all the values y2 k1 k2 over 2 so this will come out to be 0 .2158 over air now this was third stream now we find we put x3 comma y3 in place of x2 comma y2 again we find k1 which is hf now this time it will be x3 comma y3 this will be 0 .2 multiply by f of 0 .6 .0 .158 so this will come out to be 0 .1632 over here and we have k2 which will be hf x3 plus h.
05:50
H .com by x3 plus k1 so here we have k2 which will be h f x3 plus k1 so here we have have hs 0 .2, f of this is 0 .8, this will be 0 .379...