00:01
In this question we are given with the differential equation dy by dx to be equals to minus 1 .2y plus 7 times e raised to power minus 0 .3x where from x is equal to 0 to x is equal to 1 .5 and with the initial condition for x as 0 we have y as equal to 3 and we are given with h that needs to be taken as 0 .5.
00:28
So using all these values here we will apply the second -order rk method and according to this first we'll solve for k1 that is equals to h time function at x0 y0 that will be equals to 0 .5 time function at 0, 3.
00:55
Here we have this as a function x, y.
00:59
So substituting the values here we get 0 .5 times here we get the result as 3 .4 so this will be equals to 1 .7.
01:08
Next we will solve for k2 that will be equal to h time function at x0 plus h, y0 plus k1 and this will be equals to 0 .5 time function at 0 .5, 4 .7 and substituting the values here we get the result to be 0 .5 times 0 .385 so this will be equals to 0 .1925.
01:36
Next we will solve for y1 that will be equals to y0 plus k1 plus k2 by 2 that is equals to 3 plus 0 .9462 and solving this we get y at 0 .5 to be equals to 3 .9462.
01:55
Next we will solve for y2 and for that we will now replace x0 y0 by x1 y1 and hence we get k1 to be equals to h time the function value at x1 y1 that will be equals to 0 .5 times the function value at 0 .5, 3 .9462 and this will be equals to 0 .5 times 1 .2895 that will be equals to 0 .6447.
02:34
Next we will solve for k2 and this will be equals to h time the function value at x1 plus h, y1 plus k1 and this will be equals to 0 .5 times the function value at 1, 4 .591 and this will be equals to 0 .5 times minus 0 .3234 and solving this further we get the result for k2 that it will be equals to minus 0 .1617...