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# hw10_p04_answer - x Y1(x iter 1 2 3 4 0.00000 0.10000...

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Problem 4. problem 28.8, Chapra & Canale 5th Ed. p779, BVP with a shooting method. For the linear differential equation that can be solved by a simple finite differences, we applied a shooting method, which are originally for nonlinear differential equations. The two results match very well. h = 0.50000 Tolerance = 0.0001000 y( 0.00000)= 0.10000, y( 4.00000)= 0.00000 1 st guess = -0.20000, 2 nd guess = -0.10000
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Unformatted text preview: x Y1(x) iter 1 2 3 4 0.00000 0.10000 0.10000 0.10000 0.10000 0.50000 0.03067 0.08762 0.04059 0.04059 1.00000 -0.01221 0.15243 0.01648 0.01648 1.50000 -0.06625 0.35231 0.00669 0.00669 2.00000 -0.17921 0.86481 0.00271 0.00271 2.50000 -0.45131 2.14490 0.00110 0.00110 3.00000 -1.12402 5.32886 0.00044 0.00044 3.50000 -2.79446 13.24289 0.00015 0.00015 4.00000 -6.94535 32.91174 0.00000 -0.00000...
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