Solve dydx y1x yx02 using milnes predictor corrector method...

Solve (dy)/(dx) = - (y)/(1+x); y(x=0)=2 using Milne's predictor-corrector method. Solve for x=0 to 10 and a step size of 2.5. Employ Runge Kutta 2nd order method to predict the startup values

Transcript

00:01 So here as you can see we have given an equation d .y divided by d .x is equal to xy plus x square where y0 is equal to 1 at x is equal to 0 .1 0 .2 and 0 .3 we will use taylor's series and we'll continue the solution at x is equals to 0 .4 by using millip predictor corrector.

00:22 So first we have given an equation y -dice which is equal to xy plus xxxx square and value of y 0 is equals to 1 so first we'll differentiate this we'll get the value xy dash plus y plus 2x and here we'll get y -dice of 0 y double -dus of 0 is so now we'll again differentiate y double -dus then we will get the value xy double dash plus y -dice plus y -dice plus y -dase plus two that means x y double -dice plus two and while differentiating this we will get the value two so now using taylor series taylor series is y is equal to y0 plus x minus x 0 y dash of 0 plus x minus x 0 plus x minus x 0 whole square divided by 2 factorial times y double dash of 0 plus x minus x0 whole cube divided by 3 factorial y triple dash of 0 plus so on so after putting x 0 is equal to 0 and y 0 is equal to 1 we will get the value of y is equal to 1 plus x minus 0 times 0 plus x minus 0 whole square divided by 2 factorial times 1 plus x minus 0 whole cube divided by 3 factorial times 2 plus so on so from here we will get the value of y is equal 2 1 plus x square divided by 2 plus x squared divided by plus da da da da so at x equals to 0 .1 value of 5 will be 1 plus 0 .1 whole square plus 0 .1 whole cube divided by 3 plus that day that dash so from here we will get the value of x at 0 .1 at 0 .1 the y is equal to 1 .053 so by using this we will also get at x equals to 0 .2 the value of y will be 1 .0267 and at x is equal to 0 .3 the value of y will be 1 .0267 and at x is equals to 0 .3 the value of y will be 1 .054 now, we will continue this at x is equal to 0 .4 by using millenay predictor character method.

05:14 So we have given, already given, the equation dy over d x is equals to xy plus x square at y value of y0 is equal to 1.

05:31 So, the starting value of millen method with x is equal to 0 .1.

05:49 So, value of x0 is equal to 0, then y0 is equal to 1, 2, so t0 is equal to y0 is equals to 0 so x1 is already given 0 .1 the value of y1 we have already find is 1 .0053 at t 1 will be 0 .11053 same x2 is x2 is close to 0 .2 and x3 is equals to 0 .3 the value of y2 will be 1 .0 2 will be 1 .0 2 to 6 7 and value of y3 will be 1 .054 and t2 will be 0 .2444...

Question

Solve (dy)/(dx) = - (y)/(1+x); y(x=0)=2 using Milne's predictor-corrector method. Solve for x=0 to 10 and a step size of 2.5. Employ Runge Kutta 2nd order method to predict the startup values

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