I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by C h n , where C is a generally unknown constant and n is the order. For Euler's method, plot the points: ( 0.1 , ϕ ( 1 ) − y 10 ) , ( .05 , ϕ ( 1 ) − y 20 ) , ( .025 , ϕ ( 1 ) − y 40 ) , ( .0125 , ϕ ( 1 ) − y 80 ) , ( .00625 , ϕ ( 1 ) − y 160 )And then fit a line to the above data of the form C h. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?Any clarification is appreciated.ThanksEdit:The LODE I am given is y ′ = x + 2 y , y ( 0 ) = 1 and the exact solution I found was ϕ ( x ) = 1 4 ( − 2 x + 5 e 2 x − 1 ).

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I am approximating a solution to a first order LODE using Euler&#039;s method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:It states that the order of numerical methods (like Euler&#039;s) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by   C      h    n  , where   C is a generally unknown constant and   n is the order. For Euler&#039;s method, plot the points:  (  0.1  ,  ϕ  (  1  )  −      y          10        )  ,  (  .05  ,  ϕ  (  1  )  −      y          20        )  ,  (  .025  ,  ϕ  (  1  )  −      y          40        )  ,  (  .0125  ,  ϕ  (  1  )  −      y          80        )  ,  (  .00625  ,  ϕ  (  1  )  −      y          160        )And then fit a line to the above data of the form   C  h. I don&#039;t understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?Any clarification is appreciated.ThanksEdit:The LODE I am given is       y    ′    =  x  +  2  y  ,  y  (  0  )  =  1 and the exact solution I found was   ϕ  (  x  )  =      1    4    (  −  2  x  +  5      e          2      x        −  1  ).

Gianna Travis · Accepted Answer

From the way I see it, if   ϕ is the exact solution, the teacher probably wants you to compare the errors at   x  =  1 for different step sizes.Take for example the one before the last  (  0.0125  ,  ϕ  (  1  )  −      y          80        )the step size to use here is h such that   n  =  80 and   0  +  n  h  =  1    ⟹    h  =  1      /    80  =  0.0125 (I didn&#039;t see that coming).The first number in the bracket is the step size to use for each point. For each point evaluate the approximated value of   ϕ  (  1  )  ≈      y    n   Then ploy the graph of points   (      h    n    ,      ϵ    n    ) where       ϵ    n    =  ϕ  (  1  )  −      y    n   and       h    n   is the equivalent step size.

I am approximating a solution to a first order LODE using Euler's method. I made two tables, one usi

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