4 apply the runge kutta method to find the approximate value of y for x 02 in steps of 01 if y y

Step 1: The Runge-Kutta method is a numerical method for approximating the solution of a differe

4 apply the runge kutta method to find the approximate value...

Transcript

00:01 So we have to find the value of y at x equal to 0 .4 so first we have been given the value of h is equal to 0 .2 and the value of y0 is equal to 1 so we have been given the function that is f of xi comma yi is equal to dy over dx is equal to minus 2 x cube plus 4 x square minus 20 x plus 3 and the value of y at 0 .4 has to be found out so what we'll do is we know the runge -kutta method where x0 is equal to 0 and the value of y0 is equal to 1 so now the value of the variable k1 is equal to h multiplied by f of xn comma yn which is equal to 0 .2 multiplied by 3 is equal to 0 .6 similarly the value of k2 is equal to h multiplied by f of xn plus h over 2 comma yn plus k1 over 2 so here in our case it becomes 1 .038 multiplied by 0 .2 similarly the value of k3 becomes h multiplied by f of xn plus h over 2 comma yn plus k2 over 2 so this becomes this we will find it in later stages so the value of k4 turns out to be h multiplied by f of xn plus h comma yn plus the value of k3 so now the value of k turns out to be 1 over 6 multiplied by k1 plus twice k2 plus twice k3 plus k4 so in this case we will find the value of y1 to be equal to y0 plus the value of k so here we have the values x0 is equal to 0 y0 is equal to 1 the value of h it has been given to 0 .2 and the value of f of x0 comma y0 is equal to 3 so now what we'll do what we have to find is the value of x1 is equal to 0 .2 at that point the value of y1 has to be found out similarly when the value of x2 is equal to 0 .4 the value of y2 have to be found out so what we'll do is when n is equal to 0 what we can say is the value of k1 to be equal to h multiplied by f of x0 comma y0 which turns out to be h multiplied by the value of function that is minus 2x0 cube plus 4 x0 square minus 20 x0 plus 3 which turns out to be 0 .2 multiplied by 3 which is equal to 0 .6 similarly the value of k2 can be found out to be using 0 .2 sorry that is h multiplied by f of x0 plus h over 2 comma y0 plus k1 over 2 so this turns out to be 0 .2 multiplied by f of 0 .2 over 2 after simplification we get 0 .2076 same that the value of k3 turns out to be 0 .2 multiplied by f of 0 .1 comma 1 .1038 which is equal to 0 .2076 the value of k4 turns out to be minus of 0 .1712 using the above formula or else we can say that value of k4 also equals to h of x0 plus h comma y0 plus k3 after simplification we get this…

Question

4- Apply the Runge-Kutta method to find the approximate value of y for x = 0.2, in steps of 0.1, if y' = y² + x, y =1 where x =0.

Question

4- Apply the Runge-Kutta method to find the approximate value of y for x = 0.2, in steps of 0.1, if y' = y² + x, y =1 where x =0.

Please give Ace some feedback