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December 5th, 2013, 11:04 PM   #1
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Solving Diff Eqns using Euler's and Mid-point formulas

the function y(x) satisfies the differential equation dy/dx=cos(xy) and the condition y(1)=2.

a) Verify that the Euler Formula with step length 0.25 gives y(1.25)=1.89596 to 5 d.p.

b) Use the mid-point formula with h=0.25 to obtain an estimate of the value of y(2). Giving answer to 3 d.p.

OK I thought I had this sorted out because with previous questions I had no problem but my answer here is different than the answer my sheet tells me I should be getting, I'll show my working.

a) Euler's Formula> yr+1=yr+hf(xr,yr)

I know that h=0.25 and that x0=1 and y0=2 I also know that f(xr,yr)={cos(1)(2)}

So for (x1,y1) I've done

y1=2+0.25{cos(1)(2)}

y1=1.895963291 or 1.89596 to 5 d.p. so part a) is done no problem

And I now have a second set of co-ordinates that are (x1,y1) they are (1.25 , 1.895963291) so I can now use the Mid-point formula to find (x2,y2)

b) Mid-point formula> yr+1=yr-1+2hf(xr,yr)

y2=y0+2hf(x1,y1)

I know that y0=2 and that x1=1.25 and y1=1.895963291, also that h=0.25

y2=2+2(0.25){cos(1.25)(1.895963291)}

y2=2.298919812

And I now I have a third set of co-ordinates that are (x2,y2) they are (1.5 , 2.298919812)

y3=y1+2hf(x2,y2)

I know that y1=1.895963291 and that x2=1.5 and y2=2.298919812, also that h=0.25

y3=1.895963291+2(0.25){cos(1.5)(2.298919812)}

y3=1.977272868

And I now have a fourth set of co-ordinates that are (x3,y3) they are (1.75 , 1.97727286

y4=y2+2hf(x3,y3)

I know that y2=2.298919812 and that x3=1.75 and y3=1.977272868, also that h=0.25

y4=2.298919812+2(o.25){cos(1.75)(1.97727286}

y4=2.122699267

And I now have a fifth set of co-ordinates that are (x4,y4) they are (2 , 2.122699267)

Now the question asks me to find the value of y(2) and I thought that is what I had done here with y(2)=2.122699267, but my answer sheet tells me that my answer should be 1.204 to 3d.p.

Any help would be much appreciated as the previous questions that I had done my answer was correct, thanks in advance
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