2.06.MoreTools - :y@xD Accumulation Publisher: Math...

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Accumulation Authors: Bill Davis, Horacio Porta and Jerry Uhl ©1996-2007 Publisher: Math Everywhere, Inc. Version 6.0 2.06 More Tools and Measurements: Techniques for Calculating Integrals BASICS B.1) Separating the variables and integrating to get formulas for solutions of some differential equations · B.1.a) Here is Mathematica 's formula for the solution of the differential equation y' @ x D = x H y @ x D + 4 L 2 with y @ 3 D = - 9. Clear @ y, x, sol D ; dsol = DSolve A9 y' @ x D == x H y @ x D + 4 L 2 , y @ 3 D == - 9 = , y @ x D , x E :: y @ x D Ø - 2 I - 81 + 10 x 2 M - 43 + 5 x 2 >> Use the technique of separating and integrating to explain where this formula comes from. · Answer: Rewrite y' @ x D = x H y @ x D + 4 L 2 to get y' @ x D H y @ x D + 4 L 2 = x. Some folks call this technique of putting all the y @ x D terms on one side and the x terms on the other side by the name "separation of variables." Integrate both sides from 3 to x to get Ÿ 3 x y' @ t D H y @ t D + 4 L 2 t = Ÿ 3 x t t , remembering that y @ 3 D = - 9: Clear @ x, y, t D ; left = 3 x y' @ t D H y @ t D + 4 L 2 t ê . y @ 3 D Ø - 9 - 1 5 - 1 4 + y @ x D This is Mathematica 's way of telling you that Ÿ 3 x y' @ t D H y @ t D + 4 L 2 t with y @ 3 D = - 9 is: betterleft = 1 13 - 1 4 + y @ x D 1 13 - 1 4 + y @ x D Now integrate the right hand side: right = 3 x t t - 9 2 + x 2 2 Here comes the formula for y @ x D : Solve @ betterleft == right, y @ x DD :: y @ x D Ø - 2 I - 225 + 26 x 2 M - 119 + 13 x 2 >> Compare: dsol :: y @ x D Ø - 2 I - 81 + 10 x 2 M - 43 + 5 x 2 >> Not bad, eh? · B.1.b) Here is Mathematica 's formula for the solution of the logistic differential equation y' @ x D = a y @ x D J 1 - y @ x D b N with y @ 0 D = 5. Clear @ y, x, a, b, c, sol D ; dsol = DSolve B: y' @ x D == a y @ x D 1 - y @ x D b , y @ 0 D == 5 > , y @ x D , x F :: y @ x D Ø 5 b a x - 5 + b + 5 a x >> Use the technique of separating and integrating to explain where this formula comes from. · Answer: Rewrite y' @ x D = a y @ x D J 1 - y @ x D b N by separating the variables: y' @ x D y @ x D I 1 - y @ x D b M = a. Integrate both sides from 0 to x to get Ÿ 0 x y' @ t D y @ t D I 1 - y @ t D b M t = Ÿ 0 x a t, remembering that y @ 0 D = 5: Clear @ x, y, t D ; left = 0 x y' @ t D y @ t D I 1 - y @ t D b M t ê . y @ 0 D Ø 5 - Log @ 5 D + Log @ 5 - b D + Log @ y @ x DD - Log @ - b + y @ x DD This is Mathematica 's way of telling you that Ÿ 0 x y' @ t D y @ t D I 1 - y @ t D b M t with y @ 0 D = 5 is - Log @ 5 D + Log @ 5 - b D + Log @ y @ x DD - Log @ - b + y @ x DD : Put this in better form, remembering that Log @ p D + Log @ q D = Log @ p q D and - Log @ r D = Log A 1 r E : Clear @ p, q, r, s D ; logrules = : Log @ p _ D + Log @ q _ D Ø Log @ p q D , - Log @ r _ D Ø Log B 1 r F> ; betterleft = left êê . logrules - Log @ 5 D + Log B H 5 - b L y @ x D - b + y @ x D F Now integrate the right hand side: right = 0 x a t a x Now set both sides equal to each other and solve for
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2.06.MoreTools - :y@xD Accumulation Publisher: Math...

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