I'm trying to solve this 1st order ODE numerically by bringing it into an explicit form, but I don't think it is valid because of the dependency on x_n in the final expression. d y d x + x = 0 d y d x = − x Alpha is the angle from point with index (n) to point with index (n+1). t a n ( α ) = d y d x t a n ( α n ) = − x n I call h the step size. t a n ( α n ) = y n − y n − 1 x n − x n − 1 t a n ( α n ) = y n − y n − 1 h When I rearrange this I obtain the following form. y n = y n − 1 + h ⋅ t a n ( α n ) y n = y n − 1 + h ⋅ ( − x n ) y n = y n − 1 − h ⋅ x n Is this the final numerical solution of this 1st order ODE? y n = y n − 1 − h ⋅ x n EDIT: I've brought it to a better form I think. y 1 = y 0 − h ⋅ x 1 y 2 = y 1 − h ⋅ x 2 = y 0 − h ⋅ x 1 − h ⋅ x 2 y 3 = y 2 − h ⋅ x 3 = y 0 − h ⋅ x 1 − h ⋅ x 2 − h ⋅ x 3 So that the final solution is the following. y n = y 0 − h ⋅ ( x 1 + . . . + x n )

Question

I&#039;m trying to solve this 1st order ODE numerically by bringing it into an explicit form, but I don&#039;t think it is valid because of the dependency on x_n in the final expression.            d      y              d      x        +  x  =  0              d      y              d      x        =  −  x  Alpha is the angle from point with index (n) to point with index (n+1).  t  a  n  (  α  )  =            d      y              d      x          t  a  n  (      α          n        )  =  −      x          n        I call h the step size.  t  a  n  (      α          n        )  =                    y                  n                    −              y                  n          −          1                                    x                  n                    −              x                  n          −          1                        t  a  n  (      α          n        )  =                    y                  n                    −              y                  n          −          1                      h    When I rearrange this I obtain the following form.      y          n        =      y          n      −      1        +  h  ⋅  t  a  n  (      α          n        )        y          n        =      y          n      −      1        +  h  ⋅  (  −      x          n        )        y          n        =      y          n      −      1        −  h  ⋅      x          n      Is this the final numerical solution of this 1st order ODE?      y          n        =      y          n      −      1        −  h  ⋅      x          n        EDIT: I&#039;ve brought it to a better form I think.      y          1        =      y          0        −  h  ⋅      x          1              y          2        =      y          1        −  h  ⋅      x          2        =      y          0        −  h  ⋅      x          1        −  h  ⋅      x          2              y          3        =      y          2        −  h  ⋅      x          3        =      y          0        −  h  ⋅      x          1        −  h  ⋅      x          2        −  h  ⋅      x          3        So that the final solution is the following.      y          n        =      y          0        −  h  ⋅  (      x          1        +  .  .  .  +      x          n        )

ontzeidena8a · Accepted Answer

With your choice of assigning a time to the slope you get the implicit Euler method for a differential equation       y    ′    (  x  )  =  f  (  x  ,  y  (  x  )  ),      y    n    =      y          n      −      1        +  h  ·  f  (      x    n    ,      y    n    )The more basic explicit Euler method has the formula      y    n    =      y          n      −      1        +  h  ·  f  (      x          n      −      1        ,      y          n      −      1        )so that the next value is computed from the already known value       y          n      −      1

mxty42ued · Accepted Answer

So in this case       y    n    =      y    0    −  h  ∗      ∑          i      =      1        n    (      x    i    )

I'm trying to solve this 1st order ODE numerically by bringing it into an explicit form, but I don't think it is valid because of the dependency on x_n in the final expression. dy/dx+x=0 dy/dx=−x

Answered question

Answer & Explanation

New Questions in Integral Calculus