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Use the Euler method?

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2 Answers

Euler's method:
 
yn+1=yn+h y'(tn,yn)
 
In your case:
 
yn+1=yn+ 0.05 (3+tn-yn)
 
Therefore,
 
y1=y(0.05)=1+ 0.05(3+0-1) =1.1
y2=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