1. Create a Matlab script that uses Euler's method to solve...

1. Create a Matlab script that uses Euler's method to solve differential equations Set up your code to specify a final time T and N time values by using linspace to define N equally spaced points on the interval [0,T]. Use a for loop to calculate the terms in a sequence You may use the following scaffold to aid you 1 2 3 4 % Set initial condition, number of terms in sequence, 5 % delta t, function handle for f(y,t). 6 7 8 % set up a storage vector for "a" AND "t" 9 10 % for loop 11 12- for 13 % Calculate next terms in sequences 14 15- end 16

Transcript

00:01 Hello everyone so this is the question that we have part a we have to use the euler's method part b we need to use eun's method and part c we need to use the midpoint method to solve the expression d by d x is equal to 1 plus x under the root of y y of 0 is equal to 1 with the value of h is equal to 0 .5 from x is equal to 0 to 1.

00:39 Now let's jump on to the solution of the question.

00:44 By euler's method first, so this is the equation.

00:49 So i would say that y of n plus 1 is equal to y of n plus h of f x n.

00:59 How do we give the euler's formula? euler's formula is nothing but y of i plus 1 is equal to y of i plus f of x i y i h so here x i is equal to 0 plus i h so y of 1 is going to be y of 0 .5 so we have the given constraints as i is equal to 0 x not is equal to 0 and h is equal to 0 .5 so my y 1 would be equal to y0 plus f 0 comma y 0 0 .5 so when we solve this this is going to be 1 plus 1 plus 0 under the root of 1 multiplied by 0 .5 which is nothing but 1 .5.

02:09 Similarly for i is equal to 1 x1 is equal to 0 .5 and h is equal to 0 .5.

02:20 If you're going to have the value of y2 is equal to y1 plus 1 plus x1 root y1 multiplied by 0 .5 so this is going to be equal to nothing but 2 .2.

02:37 4186 and so on and so forth so that x is equal to 0 .5 the value of y is 2 .4186 now coming on to part b of the question so the predictor equation is given by or the hewn's equation is given by y of i plus 1 is equal to y of i plus f of x i y i plus f of x i y i plus f of x i plus f of x i plus 1 comma y i plus 1 which divided by so at i is equal to 0 h is equal to 0 .5 so when we put this value so y 1 of 0 would be y0 plus 1 plus x0 under the root of y 0 so this is coming up to be y1 0 is equal to 1 .5 similarly, the second expression, you're going to have it as take i is equal to 0, h is equal to 0 .5.

03:52 So my y1 would be equal to 1 plus 1 plus 0 under the root of 1 plus 1 plus 0 .5 under the root of 1 .5, multiplied by 0 .5, divided by 2.

04:07 So, y1 is coming out to be equal to 1 .709...

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