_{n+1}=y

_{n}+h y'(t

_{n},y

_{n})

_{n+1}=y

_{n}+ 0.05 (3+t

_{n}-y

_{n})

_{1}=y(0.05)=1+ 0.05(3+0-1) =1.1

_{2}=y(0.10)=1.1+0.05(3+0.05-1.1) = 1.1975

Use the Euler method with h=0.05 to find approximate values of y'=3+t-y, y(0)=1 at t=0.1, 0.2, 0.3, and 0.4.

Answer: 1.1975, 1.38549, 1.56491, 1.73658

Tutors, sign in to answer this question.

Euler's method:

y_{n+1}=y_{n}+h y'(t_{n},y_{n})

In your case:

y_{n+1}=y_{n}+ 0.05 (3+t_{n}-y_{n})

Therefore,

y_{1}=y(0.05)=1+ 0.05(3+0-1) =1.1

y_{2}=y(0.10)=1.1+0.05(3+0.05-1.1) = 1.1975

etc.

It's easiest to use a programming language to code up this method with a recursive loop.

y(n+1) = y'(n)h + y(n), where y'(n) = 3+t(n)-y(n)

y(0) = 1

y(0.05) = 1.1

y(0.1) = 1.1975

...

y(0.2) = 1.38549375

...

y(0.3) = 1.564988109

...

y(0.4) = 1.736579569

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