Consider the initial value problem given below. (dx)/(dt) =...

Consider the initial value problem given below. \frac{dx}{dt} = 1 + t \sin(tx), x(0) = 0 Use the improved Euler's method with tolerance to approximate the solution to this initial value problem at t = 1.1. For a tolerance of \epsilon = 0.02, use a stopping procedure based on absolute error. The approximate solution is x(1.1) \approx \boxed{} (Round to three decimal places as needed.)

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00:02 Hello, here in this question we have given that the in the given problem, in the given problem, we have dx by dt is equal to 2 plus t sine tx.

00:28 And we know that given that x 0 is equal to 0.

00:33 Let's say this is our equation number 1.

00:36 Now let let h equal to 13 by 40 and which is equal to 0 .32.

00:54 Now using improved iller method, using improved using improved i learned method, we have x n plus 1 is equal to x n plus 1 is plus h by 2 f t n x n plus f t n plus f t n plus 1 x bar n plus 1 now a value of value of x bar n plus 1 is we can say that it is equal to x n plus n plus h dot f t n x n x n and and the value of the value of t tn plus 1 t n plus 1 this is equal to tn plus h which is equal to tn plus h now now we have we have h is equal to 13 by 40 and t0 is equal to 0 x not so f t x will become equal to the f t x t x plus 2 now what we have to do let's suppose this is our step one let step one step one we have t1 is equal t0 plus h which is equal to 0 plus 30 by 40 which is equal to 13 by 40 so we have to solve for this so we get x x1 bar which is equal to x0 plus h f t0 x0 which is equal to 0 plus h and the value of f t not and h not which is equal to 0 which is equal to f0 f0 0 0 so from here we get 0 plus 13 by 40 into 2 which is equal to 13 by 20 now so from here we get x1 which is equal to x0 plus h by 2 f t0 x0 t not x0 t not x0 plus f t1 x1 bar on solving this we get putting the value and solving we get 0 .6610 1070 1073 845 5 1 now let us come to the equation number 2 let us come to the question number 2 let us come step 2 now what we take we take t 2 which is equal to t1 plus h t1 plus h so the value of t1 we know that it is 13 by 40 and the h equal to 13 by 40 which will become equal to the 13 by 20 so from here on solving this we get the value of x2 and on similar to item we have to solve so we get the value of x2 is 1 .333 3 5 .1.

05:34 Similarly, in step 3, on solving, we get the value of x3.

05:43 Now, we have to put, what we have to put, x2 equal to x1 plus h by 2 by iler's equation f t1, ft1, ft1, x1, x1, from here we get the value of x2, this is on x2 bar.

06:21 So we get the value of x2 is equal to 1 .40284...

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