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(Solved): y_(i+1)=y_(i)+[(1)/(3)k1+(2)/(3)k2]h k1=f(\xi ,yi) k2=f(\xi +(3)/(4)h,yi+(3)/(4)k1h) (5 Marks) (b) S ...



y_(i+1)=y_(i)+[(1)/(3)k1+(2)/(3)k2]h k1=f(\xi ,yi) k2=f(\xi +(3)/(4)h,yi+(3)/(4)k1h)

(5 Marks) (b) Solve the above function using Laplace transform method. Construct a table to compare

x

and

y

values at each step with step size

h

. (5 marks)



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