Use the two second-order multi-step methods ω i + 1 = ω i + h 2 ( 3 f i − f i − 1 )and ω i + 1 = ω i + h 2 ( f i + 1 + f i )as a pblackictor-corrector method to compute an approximation to y ( 0.3 ), with stepsize h = 0.1, for the IVP; y ′ ( t ) = 3 t y , y ( 0 ) = − 1.Use Euler’s method to start.I do not understand how to use these methods to approximate y ( 0.3 ). Moreover I am not sure how Euler's method fits into this question. Could someone clarify this question please?

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Use the two second-order multi-step methods            ω              i      +      1        =            ω        i    +      h    2    (  3      f    i    −      f          i      −      1        )and            ω              i      +      1        =            ω        i    +      h    2    (      f          i      +      1        +      f    i    )as a pblackictor-corrector method to compute an approximation to   y  (  0.3  ), with stepsize   h  =  0.1, for the IVP;      y    ′    (  t  )  =  3  t  y  ,  y  (  0  )  =  −  1.Use Euler’s method to start.I do not understand how to use these methods to approximate   y  (  0.3  ). Moreover I am not sure how Euler&#039;s method fits into this question. Could someone clarify this question please?

Madilyn Dunn · Accepted Answer

The      ω          i      +      1        =      ω    i    +      h    2    (  3  f  (      t    i    ,      ω    i    )  −  f  (      t          i      −      1        ,      ω          i      −      1        )  )is an explicit two-step (using three points   i  -  1,  i and   i  +  1) method. In contrast, the second one      ω          i      +      1        =      ω    i    +      h    2    (  f  (      t          i      +      1        ,      ω          i      +      1        )  +  f  (      t          i        ,      ω    i    )  )is an implicit one-step (using two points   i and   i  +  1) method. It is implicit since you can not easily solve it for       ω          i      +      1      , you have an equation for it      ω          i      +      1        −      h    2    f  (      t          i      −      1        ,      ω          i      +      1        )  =      ω          i        +      h    2    f  (      t          i        ,      ω          i        )  .Instead of solving that equation you&#039;re advised to use a pblackictor-corrector scheme, i.e. use       ω          i      +      1      , obtained from the first scheme and plug it into   f  (      ω          i      +      1        ) term of the second one. I&#039;ll rewrite it using tilde notation (                    ω        ~                    i      +      1       is a pblackicted value, while       ω          i      +      1       is a corrected one):  Pblackictor:                     ω        ~                    i      +      1        =      ω    i    +      h    2    (  3  f  (      t          i        ,      ω    i    )  −  f  (      t          i      +      1        ,      ω          i      −      1        )  )    Corrector:       ω          i      +      1        =      ω    i    +      h    2    (  f  (      t          i      +      1        ,                    ω        ~                    i      +      1        )  +  f  (      t          i        ,      ω    i    )  )  .Now the evaluation is really straitforward, you just take       ω          i      −      1        ,      ω          i      , compute                     ω        ~                    i      +      1       and finally       ω          i      +      1      . The only problem is that you cannot start with that. The only initial value you have is       ω    0    =  −  1 which is not sufficient to compute       ω    1   using pblackictor-corrector pair (it would need also       ω          −      1      ). To overcome this problem you can use some other method to compute       ω    1  , explicit Euler for example:      ω    1    =      ω    0    +  h  f  (      t    0    ,      ω    0    )  .

Use the two second-order multi-step methods omega_(i+1)=omega_i+h/2(3f_i−f_(i−1)) and omega_(i+1)=omega_i+h/2(f_i+1+f_i) as a pblackictor-corrector method to compute an approximation to y(0.3), with stepsize h=0.1, for the IVP; y′(t)=3ty,y(0)=−1. Use Euler’s method to start.

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