Tema26 - CHAPTER 26 26.1 (a) h < 2/100,000 = 2105. (b)...

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CHAPTER 26 26.1 ( a ) h < 2/100,000 = 2 × 10 - 5 . ( b ) The implicit Euler can be written for this problem as ( 29 y y y e e h i i i x x i i + + - - = + - + - + + 1 1 100 000 100 000 1 1 , , which can be solved for y y e h e h h i i x x i i + - - = + - + + + 1 100 000 1 100 000 1 1 , , The results of applying this formula for the first few steps are shown below. A plot of the entire solution is also displayed x y 0 0 0.1 1.904638 0.2 1.818731 0.3 1.740819 0.4 1.67032 0.5 1.606531 0 1 2 0 1 2 26.2 The implicit Euler can be written for this problem as ( 29 y y t y t h i i i i i + + + + = + - + 1 1 1 1 30 3 (sin ) cos which can be solved for y y t h t h h i i i i + + + = + + + 1 1 1 30 3 1 30 sin cos The results of applying this formula are tabulated and graphed below. x y x y x y x y 0 0 1.2 0.952306 2.4 0.622925 3.6 -0.50089 0.4 0.444484 1.6 0.993242 2.8 0.270163 4 -0.79745 0.8 0.760677 2 0.877341 3.2 -0.12525
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-1 -0.5 0 0.5 1 0 1 2 3 4 26.3 ( a ) The explicit Euler can be written for this problem as ( 29 x x x x h i i i i 1 1 1 1 2 999 1999 , , , , + = + + ( 29 x x x x h i i i i 2 1 2 1 2 1000 2000 , , , , + = + - - Because the step-size is much too large for the stability requirements, the solution is unstable, t x1 x2 dx1 dx2 0 1 1 2998 -3000 0.05 150.9 -149 -147102 147100 0.1 -7204.2 7206 7207803 -7207805 0.15 353186 -353184 -3.5E+08 3.53E+08 0.2 -1.7E+07 17305943 1.73E+10 -1.7E+10 ( b ) The implicit Euler can be written for this problem as ( 29 x x x x h i i i i 1 1 1 1 1 2 1 999 1999 , , , , + + + = + + ( 29 x x x x h i i i i 2 1 2 1 1 2 1 1000 2000 , , , , + + + = + - - or collecting terms ( ) ( ) , , , , , , 1 999 1999 1000 1 2000 1 1 2 1 1 1 1 2 1 2 - - = + + = + + + + h x hx x hx h x x i i i i i i or substituting h = 0.05 and expressing in matrix format - - = + + 48 95 99 95 50 101 1 1 2 1 1 2 . . , , , , x x x x i i i i Thus, to solve for the first time step, we substitute the initial conditions for the right-hand side and solve the 2x2 system of equations. The best way to do this is with LU decomposition since we will have to solve the system repeatedly. For the present case, because its easier to display, we will use the matrix inverse to obtain the solution. Thus, if the matrix is inverted, the solution for the first step amounts to the matrix multiplication, { } { } x x i i 1 1 2 1 1886088 186648 0 93371 0 9141 1 1 3752568 184781 , , . . . . . . + + = - - = - For the second step (from x = 0.05 to 0.1),
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{ } { } x x i i 1 1 2 1 1886088 186648 0 93371 0 9141 3752568 184781 362878 181472 , , . . . . . . . . + + = - - - = - The remaining steps can be implemented in a similar fashion to give t x1 x2 0 1 1 0.05 3.752568 -1.84781 0.1 3.62878 -1.81472 0.15 3.457057 -1.72938 0.2 3.292457 -1.64705 The results are plotted below, along with a solution with the explicit Euler using a step of 0.0005. -2
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Tema26 - CHAPTER 26 26.1 (a) h &lt; 2/100,000 = 2105. (b)...

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