Euler's method

  • Thread starter lmanri
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Homework Statement


Use EulersMethod to perform Euler's method with the given step size Δt on the given initial value problem over the time interval specified :

dy/dt= (y^2)-4t , y(0)=0.5 , 0<=t<=2, Δt=0.25





The Attempt at a Solution




This is what I did but I dont think its right, because in the back of the book there is a different answer. I really need to know how to work it out please help.


y1= 0.5+(.5^2-(4)(0)).25= .05625
y2= 0.5625+(.5625^2-(4)(.25)).25= .3916
y3= 0.3916+(.3916^2-(4)(.50)).25= -.07006
y4= -.07006+(-.07006^2-(4)(.75)).25= -.821289
y5= -.821289+(-.821289^2-(4)(1.0)).25=-1.98992
y6= -1.98992+(-1.98992^2-(4)(1.25)).25= -4.22987
y7= -4.22987+(-4.22987^2-(4)(1.50)).25= -10.2028
y8= -10.2028+(-10.2028^2-(4)(1.75)).25= -37.9771





The book says: y1= 0.56, y2=0.39, y8=-2.69

I don't know how they got to y8=-2.69
 

Answers and Replies

  • #2
eumyang
Homework Helper
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y1= 0.5+(.5^2-(4)(0)).25= .05625
y2= 0.5625+(.5625^2-(4)(.25)).25= .3916
y3= 0.3916+(.3916^2-(4)(.50)).25= -.07006
y4= -.07006+(-.07006^2-(4)(.75)).25= -.821289
Here's where you started going wrong. You need an extra set of parentheses:
[tex]y_4 = -.07006 + ((-.07006)^2 - 4(0.75))(0.25) = -0.81883[/tex]
Remember that (-a)2 is not the same as -a2.
 

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