Use the Euler Method to write down an iterative algorithm so that y n + 1 = y ( t n + 1 ) can be determined from y n = y ( t n ), where t n = n △ t is the size of the time step, for the following Ordinary Differential Equation d y d t = y 2 + sin ⁡ ( t )I'm confused about this question because the question's I'm been doing upto this one is that I am given an initial ( x 0 , y 0 ) and then I draw up a table to find a y value for a corresponding x value.. so this question is very confusing as I have never came across a question such as this one.

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Use the Euler Method to write down an iterative algorithm so that       y          n      +      1        =  y  (      t          n      +      1        ) can be determined from       y    n    =  y  (      t    n    ), where       t    n    =  n  △  t is the size of the time step, for the following Ordinary Differential Equation             d      y              d      t        =      y    2    +  sin  ⁡  (  t  )I&#039;m confused about this question because the question&#039;s I&#039;m been doing upto this one is that I am given an initial   (      x    0    ,      y    0    ) and then I draw up a table to find a   y value for a corresponding   x value.. so this question is very confusing as I have never came across a question such as this one.

Kailee Abbott · Accepted Answer

Euler&#039;s method for the differential equation       y    ′    =  f  (  t  ,  y  ) and step size   Δ  t gives:                    y                  n          +          1                    −              y                  n                            Δ      t        =  f  (      t    n    ,      y    n    )Where       t    n    =  n  Δ  t. This question is basically asking you to solve for       y          n      +      1       in this specific case. So you get:      y          n      +      1        =      y    n    +  Δ  t  (      y    n    2    +  sin  ⁡  (  n  Δ  t  )  )

Widersinnby7 · Accepted Answer

The way I was taught this:            Δ      y              Δ      t        ≈            d      y              d      t        =      y    2    +  sin  ⁡  (  t  )      y          n      +      1        −      y    n    =  Δ  y  ≈  Δ  t  (      y    n    2    +  sin  ⁡  (      t    n    )  )      y          n      +      1        ≈      y    n    +  Δ  t  (      y    n    2    +  sin  ⁡  (  n  Δ  t  )  )This allows you to take a given initial coordinate   (      t    0    ,      y    0    ) and   Δ  t and quickly run through points by an iterative algorithm in your calculator.

use the Euler Method to write down an iterative algorithm so that y_(n+1)=y(t_(n+1)) can be determined from y_n=y(t_n), where t_n=n triangle t is the size of the time step, for the following Ordinary Differential Equation dy/dt=y^2+sin(t)

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