00:01
Hi, today we are solving the question in which we are given with the initial value problem d, d, y, by dt is equals to y into t square minus 1 .1 into y and we are given it y not is equals to 1.
00:18
So we have to solve it analytically first.
00:21
So let us solve it analytically firstly so in order to solve it analytic firstly by using separation of variables so 1 by y into dyy is equal to t square minus 1 .1 into d t so from here taking the integral of 1 by y into dy y is equal to integral of t square minus 1 .1 into d t on solving the integral we get we get loan y is equal to t cube minus t cube divided by three minus 1 .1 into t plus c so on solving it y is equal to c into e raised to the part t cube divided by 3 minus 1 .1 into t so from here y not is equals to 1 this implies 1 is equal to c so from here we get our equation y is equals to e raised to the part t cubed divided by three minus 1 .1 t now taking the b so we have to take h is equal to 0 .5 firstly and so we get n is equal to 4 by n is equals to b minus a divided by h so by uller's method t i plus 1 is given by t i plus 1 is equal to i into h where y i i plus 1 is equals to yi plus h into y i into y i into t i whole square minus 1 .1 into yi now solving it the table will be look like so this is the required table similarly if we take h is equals to 0 .25 so this is the required table so this is the required table so through euler's method.
02:24
Now taking the part c, initially t0 is equal to 0 and y0 is equals to 1.
02:31
So f t y is equal to y t square minus 1 .1 into y.
02:40
Now from here we get from the mid point method, y i plus 1 is equal to y i plus h into f of t i plus 1 by 2 and y i plus 1 by 2.
02:54
So let us make the table so the iteration goes as follows.
03:00
So the iteration will be like this, iti and yi...