Part I A. [20 Points] Using centered finite difference approximations as done in class, solve the...

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Part I A. [20 Points] Using centered finite difference approximations as done in class, solve the equation: 1.045 d20 dx + 1.2Q = 0 subject to the boundary conditions shown in the stencil below. Do this for two values of Q: (a) Q = 0.1, and (b) Q = (1 + 2x)e-sinx(cos (2.3x) + 1.113sin2.7x + (0.333cos x + ex-sinx)x+.007x. For Q = 0.1, use the stencil in Fig. 1. For Q = (1 + 2x)e-sinx(cos (2.3x) + 1.113sin2.7x + (0.333cos x + ex-sinx)x+.007x, calculate with both the stencils in Fig. 1 and Fig 2. For all the three cases, show a table as well as a plot of e versus x. Hand in any MATLAB codes if you use MATLAB. 1 2 3 4 0=0 x=0 Fig 1 1 2 3 4 5 6 7 8 9 10 0=0 x=0 Fig 2 =1 x=1 11 0=1 x=1 Part I A. [20 Points] Using centered finite difference approximations as done in class, solve the equation: 1.045 d20 dx + 1.2Q = 0 subject to the boundary conditions shown in the stencil below. Do this for two values of Q: (a) Q = 0.1, and (b) Q = (1 + 2x)e-sinx(cos (2.3x) + 1.113sin2.7x + (0.333cos x + ex-sinx)x+.007x. For Q = 0.1, use the stencil in Fig. 1. For Q = (1 + 2x)e-sinx(cos (2.3x) + 1.113sin2.7x + (0.333cos x + ex-sinx)x+.007x, calculate with both the stencils in Fig. 1 and Fig 2. For all the three cases, show a table as well as a plot of e versus x. Hand in any MATLAB codes if you use MATLAB. 1 2 3 4 0=0 x=0 Fig 1 1 2 3 4 5 6 7 8 9 10 0=0 x=0 Fig 2 =1 x=1 11 0=1 x=1