Consider the IVP y ″ = − y for t ≥ 0, and y ( 0 ) = 1, y ′ ( 0 ) = 2.I have rewritten this differential equation as a system of first-order ODE's such that u ′ = v v ′ = − u with u ( 0 ) = 1 , v ( 0 ) = 2.The solution is y = 2 sin ⁡ ( t ) + cos ⁡ ( t ), y ′ = 2 cos ⁡ ( t ) − sin ⁡ ( t )I am asked to perform one step of Euler's method with h = 0.5, and determine if Euler's method is stable for this problem.For the first part, I find that one step of Euler's method yields y 1 = y 0 + h f ( t 0 , y 0 ) = 1 + ( 0.5 ) ( 2 cos ⁡ ( 0 ) − sin ⁡ ( 0 ) ) = 2. But how do I determine if Euler's method is stable? I know that for the equation y ′ = λ y, Euler's method is stable for | 1 + h λ | ≤ 1, but since this problem is in a different form I'm not sure what I need to do here.

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Consider the IVP                              y          ″                =        −        y            for   t  ≥  0, and   y  (  0  )  =  1,       y    ′    (  0  )  =  2.I have rewritten this differential equation as a system of first-order ODE&#039;s such that                              u          ′                =        v                                      v          ′                =        −        u            with   u  (  0  )  =  1  ,  v  (  0  )  =  2.The solution is   y  =  2  sin  ⁡  (  t  )  +  cos  ⁡  (  t  ),       y    ′    =  2  cos  ⁡  (  t  )  −  sin  ⁡  (  t  )I am asked to perform one step of Euler&#039;s method with   h  =  0.5, and determine if Euler&#039;s method is stable for this problem.For the first part, I find that one step of Euler&#039;s method yields                              y          1                =                  y          0                +        h        f        (                  t          0                ,                  y          0                )        =        1        +        (        0.5        )        (        2        cos        ⁡        (        0        )        −        sin        ⁡        (        0        )        )        =        2.            But how do I determine if Euler&#039;s method is stable? I know that for the equation       y    ′    =  λ  y, Euler&#039;s method is stable for       |    1  +  h  λ      |    ≤  1, but since this problem is in a different form I&#039;m not sure what I need to do here.

Bornejecbo · Accepted Answer

Applying the Euler iteration procedure we have      {                                        u            k                    =                      u                          k              −              1                                +          h                      v                          k              −              1                                                                        v            k                    =                      v                          k              −              1                                −          h                      u                          k              −              1                                              or      (                                        u            k                                                            v            k                                )    =      (                            1                          h                                      −          h                          1                      )        (                                        u                          k              −              1                                                                        v                          k              −              1                                            )  or      U    k    =      M    k        U    0  this sequence converges as long as the eigenvalues of   M have absolute value less than 1. Here the   M eigenvalues are   1  ±  i  h with absolute value       1    +          h      2        &amp;gt;  1 so the Euler procedure diverges.

Layla Velazquez · Accepted Answer

You apply the Euler step to the first-order system.      u    1    =      u    0    +  h      u    0    ′    =      u    0    +  h      v    0    ,        v    1    =      v    0    +  h      v    0    ′    =      v    0    −  h      u    0    .As the Euler step is tangential to the convex solution curve, it will always move outwards, away from the center of the concentric exact solution curves.

Consider the IVP <mtable columnalign="right left right left right left right left right left

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