In such cases dsolve may return solutions containing

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Unformatted text preview: k + 2 α cosh( ) m)Heaviside(t − 1) αm √ %1 t α m (−1 + cosh( )) αm 2 (1 + α)2 ) − 2 /m, x1 (t) = (k k (1 + α) √ %1 t √ √ α cosh( ) %1 %1 (− α m t ) ( α mt ) αm +e −2 e 1+α √ %1 (t − 1) Heaviside(t − 1) cosh( ) αm − +( k (1 + α)2 1 k Heaviside(t − 1) (1 + α) t2 2 − k Heaviside(t − 1) (1 + α) t + Heaviside(t − 1) m 1 1 + Heaviside(t − 1) k + Heaviside(t − 1) α k 2 2 x2 (t) = k + 2 α m k + 2 α2 m k ) (m k (1 + α)2 ) %1 := −α m k (1 + α) Evaluate at values for the constants. > eval( %, {alpha=1/10, m=1, k=1} ); 7.2 Ordinary Differential Equations • 249 {x2 (t) = ( √ 9 11 2 11 1 1√ −11 100 (t − 1))) + t− t + cosh( 10 10 5 5 10 √ 2 2 1√ Heaviside(t − 1) + −11 100 t), x1 (t) − cosh( 11 11 10 √ √ √ √ (−1/10 −11 100 t) + e(1/10 −11 100 t) =e √ 2 1√ − cosh( −11 100 t) 11 10 √ 1√ 100 − Heaviside(t − 1) cosh( −11 100 (t − 1)) 121 10 10 5 + Heaviside(t − 1) t2 − Heaviside(t − 1) t 11 11 155 2 + Heaviside(t − 1) + } 121 11 As expected, you get the same solution as before. The type=series Option The series method for solving differential equations finds an approximate symbolic solution to the equations in the following manner. Maple finds a series approximation to the equations. It then solves the series approximation symbolically, using exact methods. This technique is useful when Maple standard algorithms fail, but you still want a symbolic solution rather than a purely numeric solution. The series method can often help with nonlinear and high-order ODEs. When using the series method, Maple assumes that a solution of the form ∞ 50 121 x c i=0 ai xi exists, where c is a rational number. Consider the following differential equation. > eq := 2*x*diff(y(x),x,x) + diff(y(x),x) + y(x) = 0; eq := 2 x ( d2 d y(x)) + ( y(x)) + y(x) = 0 2 dx dx Solve the equation. > dsolve( {eq}, {y(x)}, type=series ); 250 • Chapter 7: Solving Calculus Problems y(x) = _C1 √ x(1 − 1 12 13 1 x+ x− x+ x4 − 3 30 630 22680 1 x5 + O(x6 )) + _C2 1247400 1 13 1 1 (1 − x + x2 − x+ x4 − x5 + O(x6 )) 6 90 2520 113400 Use rhs to select the solution, then convert it to a polynomial. > rhs(%); 13 1 1 1 12 x+ x− x+ x4 − 3 30 630 22680 1247400 5 + O(x6 )) + _C2 x 13 1 1 1 (1 − x + x2 − x+ x4 − x5 + O(x6 )) 6 90 2520 113400 _C1 x(1 − > poly := convert(%, polynom); √ √ poly := _C1 x 1 12 13 1 1 (1 − x + x− x+ x4 − x5 ) 3 30 630 22680 1247400 1 13 1 1 x+ x4 − x5 ) + _C2 (1 − x + x2 − 6 90 2520 113400 Plot the solution for different values of the arbitrary constants _C1 and _C2. > [ seq( _C1=i, i=0..5 ) ]; [_C1 = 0, _C1 = 1, _C1 = 2, _C1 = 3, _C1 = 4, _C1 = 5] > map(subs, %, _C2=1, poly); 7.2 Ordinary Differential Equations • 251 12 13 1 1 x− x+ x4 − x5 , 6 90 2520 113400 1 13 1 1 %1 + 1 − x + x2 − x+ x4 − x5 , 6 90 2520 113400 1 13 1 1 2 %1 + 1 − x + x2 − x+ x4 − x5 , 6 90 2520 113400 13 1 1 1 x+ x4 − x5 , 3 %1 + 1 − x + x2 − 6 90 2520 113400 1 13 1 1 4 %1 + 1 − x + x2 − x+ x4 − x5 , 6 90 2520 113400 1 13 1 1 5 %1 + 1 − x + x2 − x+ x4 − x5 ] 6 90 2520 113400 %1 := √ 1 13 1 1 12 x (1 − x + x− x+ x4 − x5 ) 3 30...
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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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