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.

wendi1019gt 2022-08-14 Answered
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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Answers (1)

Madilyn Dunn
Answered 2022-08-15 Author has 16 answers
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'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'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 ) .
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