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# Here is a dierential equation and an initial

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Unformatted text preview: ln(_a ) + _C1 Use the explicit option to search for an explicit solution for the ﬁrst result. > dsolve( {de2}, {y(x)}, explicit ); {y(x) = _C1 }, _Z y(x) = RootOf − 1 d_f + x + _C2 _f ln(_f ) + _C1 However, in some cases, Maple may not be able to ﬁnd an explicit solution. There is also an implicit option to force answers to be returned in implicit form. The method=laplace Option Applying Laplace transform methods to diﬀerential equations often reduces the complexity of the problem. The transform maps the diﬀerential equations into algebraic equations, which are much easier to solve. The diﬃculty is in the transformation of the equations to the new domain, and especially the transformation of the solutions back. The Laplace transform method can handle linear ODEs of arbitrary order, and some cases of linear ODEs with non-constant coeﬃcients, provided that Maple can ﬁnd the transforms. This method can also solve systems of coupled equations. Consider the following problem in classical dynamics. Two weights with masses m and αm, respectively, rest on a frictionless plane joined by a spring with spring constant k . What are the trajectories of each weight if the ﬁrst weight is subject to a unit step force u(t) at time t = 1? First, set up the diﬀerential equations that govern the system. Newton’s Second Law governs the motion of the ﬁrst weight, and hence, the mass m times the acceleration must equal the sum of the forces that you apply to the ﬁrst weight, including the external force u(t). > eqn1 := > alpha*m*diff(x[1](t),t\$2) = k*(x[2](t) - x[1](t)) + u(t); eqn1 := α m ( d2 x1 (t)) = k (x2 (t) − x1 (t)) + u(t) dt2 244 • Chapter 7: Solving Calculus Problems Similarly for the second weight. > eqn2 := m*diff(x[2](t),t\$2) = k*(x[1](t) - x[2](t)); eqn2 := m ( d2 x2 (t)) = k (x1 (t) − x2 (t)) dt2 Apply a unit step force to the ﬁrst weight at t = 1. > u := t -> Heaviside(t-1); u := t → Heaviside(t − 1) At time t = 0, both masses are at rest at their respective locations. > ini := x[1](0) = 2, D(x[1])(0) = 0, > x[2](0) = 0, D(x[2])(0) = 0 ; ini := x1 (0) = 2, D(x1 )(0) = 0, x2 (0) = 0, D(x2 )(0) = 0 Solve the problem using Laplace transform methods. > dsolve( {eqn1, eqn2, ini}, {x[1](t), x[2](t)}, > method=laplace ); 1 x2 (t) = (−2 α m + t2 k + t2 k α − 2 t k − 2 t k α + k + α k 2 √ %1 (t − 1) ) m)Heaviside(t − 1) (k + 2 α cosh( αm √ %1 t α (−1 + cosh( )) 1 αm , x1 (t) = (2 m (1 + α)2 m) − 2 1+α 2 √ %1 (t − 1) − 2 cosh( ) m + t2 k + t2 k α − 2 t k − 2 t k α αm + k + α k )Heaviside(t − 1) (k m (1 + α)2 ) + ( √ √ √ √ %1 t ) αm e(− + e(− α m ) α + e( α m √ %1 t − 2 α cosh( ))/(1 + α) αm %1 t ) αm %1 t %1 t ) + e( α + 2α %1 := −α m k (1 + α) 7.2 Ordinary Diﬀerential Equations • 245 Evaluate the result at values for the constants. > ans := eval( %, {alpha=1/10, m=1, k=1} ); ans := {x2 (t) = ( √ 9 11 2 11 1 1√ + t− t + cosh( −11 100 (t − 1))) 10 10 5 5 10 √ 2...
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## This note was uploaded on 08/27/2012 for the course MATH 1100 taught by Professor Nil during the Spring '12 term at National University of Singapore.

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