290_pdfsam_math 54 differential equation solutions odd

# 290_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 5 13. See the answer in the text. 15. See the answer in the text. 17. Let x1 := u and x2 := v , and denote the independent variable by t (in order to be consistent with formulas in Section 5.3). In new notation, we have an initial value problem x1 = 3x1 − 4x2 , x2 = 2x1 − 3x2 , x1 (0) = x2 (0) = 1 for a system in normal form. Here f1 (t, x1 , x2 ) = 3x1 − 4x2 , f2 (t, x1 , x2 ) = 2x1 − 3x2 . Thus formulas for ki,j ’s in vectorized Runge-Kutta algorithm become k1,1 = h(3x1;n − 4x2;n ), k2,1 = h(2x1;n − 3x2;n ), k1,1 k2,1 k1,2 = h 3 x1;n + − 4 x2;n + 2 2 k1,1 k2,1 − 3 x2;n + k2,2 = h 2 x1;n + 2 2 k1,2 k2,2 − 4 x2;n + k1,3 = h 3 x1;n + 2 2 k1,2 k2,2 − 3 x2;n + k2,3 = h 2 x1;n + 2 2 k1,4 = h [3 (x1;n + k1,3 ) − 4 (x2;n + k2,3 )] , k2,4 = h [2 (x1;n + k1,3 ) − 3 (x2;n + k2,3 )] . With the inputs t0 = 0, x1;0 = x2;0 = 1, and step size h = 1 we compute k1,1 = h(3x1;0 − 4x2;0 ) = 3(1) − 4(1) = −1, k2,1 = h(2x1;0 − 3x2;0 ) = 2(1) − 3(1) = −1, k1,1 k2,1 −1 k1,2 = h 3 x1;0 + − 4 x2;0 + =3 1+ 2 2 2 286 −1 2 1 =− , 2 , , , , −4 1+ ...
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