1. Write this second order ode as a first order ode. y ″ ( t ) = y ( t ) + y ′ ( t ) + t y ′ ( 0 ) = 1 y ′ ( 0 ) = 0Use the Euler Method with h = 0.1. What approximation of ( y ( h ) , y ′ ( h ) ) do you get?"I think I understand the first problem. I substitute y with u 1 and y ′ with u 2 which gives: y ′ = u 2 , y ″ = u 1 + u 2 + t. Is that correct?If so, I don't really know where to go from there.

Question

1. Write this second order ode as a first order ode.      y    ″    (  t  )  =  y  (  t  )  +      y    ′    (  t  )  +  t      y    ′    (  0  )  =  1      y    ′    (  0  )  =  0Use the Euler Method with h = 0.1. What approximation of   (  y  (  h  )  ,      y    ′    (  h  )  ) do you get?&quot;I think I understand the first problem. I substitute y with       u    1   and       y    ′   with       u    2   which gives:       y    ′    =      u    2    ,      y    ″    =      u    1    +      u    2    +  t. Is that correct?If so, I don&#039;t really know where to go from there.

Abigayle Lynn · Accepted Answer

Expanding the comment of Winther: Yes, but write       y    ″    =      u    2    ′   to get the first order system:                              u          1          ′                                    =                  u          2                             and                                       u          2          ′                                    =                  u          1                +                  u          2                +        t        .             and Now apply Euler&#039;s method one step:                    y        (        h        )        =                  u          1                (        h        )                            ≈                  u          1                (        0        )        +        h                          u          2                (        0        )                                  =        1        +        h        ⋅        0                     and                                       y          ′                (        h        )        =                  u          2                (        h        )                            ≈                  u          2                (        0        )        +        h                                      [                                    u          1                (        0        )        +                  u          2                (        0        )        +        0                              ]                                                    =        0        +        h        ⋅        [        1        +        0        +        0        ]        .             and In the next step you would get                    y        (        2        h        )        =                  u          1                (        2        h        )                            ≈                  u          1                (        h        )        +        h                          u          2                (        h        )                                  ≈        1        +        h        ⋅        h                     and                                       y          ′                (        2        h        )        =                  u          2                (        2        h        )                            ≈                  u          2                (        h        )        +        h                                      [                                    u          1                (        h        )        +                  u          2                (        h        )        +        h                              ]                                                    ≈        h        +        h        ⋅        [        1        +        h        +        h        ]        .            and et

1. Write this second order ode as a first order ode. y′′(t)=y(t)+y′(t)+t y(0)=1 y′(0)=0 2. Use the Euler Method with h = 0.1. What approximation of (y(h),y′(h)) do you get?

Answered question

Answer & Explanation

New Questions in Integral Calculus