Use a numerical solver and Euler's method to obtain a...

Use a numerical solver and Euler's method to obtain a four-decimal approximation of the indicated value. First use $h = 0.1$ and then use $h = 0.05$. $y' = (x - y)^2$, $y(0) = 0.5$; y(0.5) $y(0.5) \approx$ \(h = 0.1) y(0.5) \approx \(h = 0.05)

Transcript

00:01 Hi, in this question we are given with the derivative of y -dash that is equals to e -d -to -par minus y and we are given with some initial condition for y -0 to be equal to 0 and y 0 .5 we need to find using euler's method to obtain a four decimal approximation for this where we have for the first part h is equal to 0 .1 now using the euler's method for h is 0 .1.

00:30 1 we get h of y of 1 to be equals to y not plus h times function value at x not y not as here we can see our x is 0 and y is 0 initially so substituting the initial value we get this to be equals to 0 plus 0 .1 times f at 0 0 will be 1 and solving this further we get this further we get this to be 0 .1 .5 .5 times f at 0 will be 1 and solving this further we get this.

01:02 This to be equal to 0 .1.

01:05 Now finding the second value y 2 at y1 plus h time function value at x1 so finding the function value over here we get this to be equals to y1 that is 0 .1 plus h is 0 .1 and the function value at 0 .1 and it will result in here we have y as 0 .1 so e is to per minus 0 .1, it will result in 0 .9048.

01:38 And solving this further, we get this to be equal to 0 .1905.

01:44 Now moving further for the next approximation y3, that would be equals to y2 plus hf at x2, here substituting the known value, we get 0 .1905 plus h .s.

01:59 0 .105 plus h .s.

02:01 0 .5 .000 1 and function value at x2 and y2 we get this as 0 .8 to double 6 and solving this we get the result as 0 .2731.

02:15 Now finding the next approximation at y3 plus h times function value at x3 y3 and this will be equals to 0 .2731 plus 0 .1 and the function value at this will result in 0 .2731 and the function value at this will result in 0 .0 .1.

02:31 0 .761 and solving this we get the result to be equal to 0 .3492.

02:38 Now finding the next approximation y5 at y4 plus h times function value at x4 y4 and here substituting the value 0 .3492 plus h as 0 .1 and the function value at these that will result in 0 .7052.

02:59 Solving this we get this to be equal to 0 .4198 and from here we get our approximation that our value at y 0 .5 will be equal to 0 .498.

03:17 Now solving further for the next part b for h is equal to 0 .05 we get the approximation y 1 for y 0 .0 plus h time function value at x0 y0.

03:31 That will result in equal to 0 plus ha 0 .05 and the function value at 0 and 0 will result in 1 and solving this we get 0 .05.

03:42 Now i'm finding the next approximation y2 at y1 as 0 .05 we get this to be equal to 0 .05 plus 0 .05 times function value at 0 .05 that will result in 0 .9512.

03:57 And solving this we get this to be equal to 0 .09...

Question

Use a numerical solver and Euler's method to obtain a four-decimal approximation of the indicated value. First use $h = 0.1$ and then use $h = 0.05$. $y' = (x - y)^2$, $y(0) = 0.5$; y(0.5) $y(0.5) \approx$ \(h = 0.1) y(0.5) \approx \(h = 0.05)

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