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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTA

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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTA
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  LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS  GRATIS EN DESCARGA DIRECTA SIGUENOS EN: VISITANOS PARA DESARGALOS GRATIS.  Table of ContentsPart I  Ordinary Differential Equations 1  Introduction to Differential Equations  12  First-Order Differential Equations  223  Higher-Order Differential Equations  994  The Laplace Transform  1985  Series Solutions of Linear Differential Equations  2526  Numerical Solutions of Ordinary Differential Equations  317 Part II  Vectors, Matrices, and Vector Calculus 7  Vectors  3398  Matrices  3739  Vector Calculus  438 Part III  Systems of Differential Equations 10  Systems of Linear Differential Equations  55111  Systems of Nonlinear Differential Equations  604 Part IV  Fourier Series and Partial Differential Equations 12  Orthogonal Functions and Fourier Series  63413  Boundary-Value Problems in Rectangular Coordinates  68014  Boundary-Value Problems in Other Coordinate Systems  75515  Integral Transform Method  79316  Numerical Solutions of Partial Differential Equations  832 www.librosysolucionarios.net  Part V  Complex Analysis 17  Functions of a Complex Variable  85418  Integration in the Complex Plane  87719  Series and Residues  89620  Conformal Mappings  919 Appendices Appendix II  Gamma function  942 Projects 3.7  Road Mirages  9443.10  The Ballistic Pendulum  9468.1  Two-Ports in Electrical Circuits  9478.2  Traffic Flow  9488.15  Temperature Dependence of Resistivity  9499.16  Minimal Surfaces  95014.3  The Hydrogen Atom  95215.4  The Uncertainity Inequality in Signal Processing  95515.4  Fraunhofer Diffraction by a Circular Aperture  95816.2  Instabilities of Numerical Methods  960 www.librosysolucionarios.net  Part I Ordinary Differential Equations 11  Introduction toDifferential Equations EXERCISES 1.1 De fi nitions and Terminology 1.  Second order; linear 2.  Third order; nonlinear because of ( dy/dx ) 4 3.  Fourth order; linear 4.  Second order; nonlinear because of cos( r  + u ) 5.  Second order; nonlinear because of ( dy/dx ) 2 or   1 + ( dy/dx ) 2 6.  Second order; nonlinear because of   R 2 7.  Third order; linear 8.  Second order; nonlinear because of ˙ x 2 9.  Writing the differential equation in the form  x ( dy/dx ) +  y 2 = 1, we see that it is nonlinear in  y  because of   y 2 .However, writing it in the form ( y 2 − 1)( dx/dy ) + x  = 0, we see that it is linear in  x . 10.  Writing the differential equation in the form  u ( dv/du ) + (1 +  u ) v  =  ue u we see that it is linear in  v . However,writing it in the form ( v  + uv − ue u )( du/dv ) + u  = 0, we see that it is nonlinear in  u . 11.  From  y  =  e − x/ 2 we obtain  y  = − 12 e − x/ 2 . Then 2 y  + y  = − e − x/ 2 + e − x/ 2 = 0. 12.  From  y  =  65  −  65 e − 20 t we obtain  dy/dt  = 24 e − 20 t , so that dydt  + 20 y  = 24 e − 20 t + 20  65  −  65 e − 20 t   = 24 . 13.  From  y  =  e 3 x cos2 x  we obtain  y  = 3 e 3 x cos2 x  −  2 e 3 x sin2 x  and  y  = 5 e 3 x cos2 x  −  12 e 3 x sin2 x , so that y  − 6 y  + 13 y  = 0. 14.  From  y  = − cos x ln(sec x + tan x ) we obtain  y  = − 1 + sin x ln(sec x + tan x ) and y  = tan x + cos x ln(sec x + tan x ). Then  y  + y  = tan x . 15.  The domain of the function, found by solving  x + 2 ≥ 0, is [ − 2 , ∞ ). From  y  = 1 + 2( x + 2) − 1 / 2 we have( y − x ) y  = ( y − x )[1 + (2( x + 2) − 1 / 2 ]=  y − x + 2( y − x )( x + 2) − 1 / 2 =  y − x + 2[ x + 4( x + 2) 1 / 2 − x ]( x + 2) − 1 / 2 =  y − x + 8( x + 2) 1 / 2 ( x + 2) − 1 / 2 =  y − x + 8 . 1 www.librosysolucionarios.net  -4 -2 2 4 t-4-224X 1.1  Definitions and Terminology An interval of definition for the solution of the differential equation is ( − 2 , ∞ ) because  y  is not defined at x  = − 2. 16.  Since tan x  is not defined for  x  =  π/ 2 +  nπ ,  n  an integer, the domain of   y  = 5tan5 x  is { x   5 x  =  π/ 2 + nπ }  or  { x   x  =  π/ 10 + nπ/ 5 } . From  y  = 25sec 2 5 x  we have y  = 25(1 + tan 2 5 x ) = 25 + 25tan 2 5 x  = 25 + y 2 . An interval of definition for the solution of the differential equation is ( − π/ 10 ,π/ 10). Another interval is( π/ 10 , 3 π/ 10), and so on. 17.  The domain of the function is  { x   4 − x 2  = 0 }  or  { x   x  = − 2 or  x  = 2 } . From  y  = 2 x/ (4 − x 2 ) 2 we have y  = 2 x   14 − x 2  2 = 2 xy. An interval of definition for the solution of the differential equation is ( − 2 , 2). Other intervals are ( −∞ , − 2)and (2 , ∞ ). 18.  The function is  y  = 1 / √  1 − sin x , whose domain is obtained from 1 − sin x  = 0 or sin x  = 1. Thus, the domainis  { x   x  =  π/ 2 + 2 nπ } . From  y  = − 12 (1 − sin x ) − 3 / 2 ( − cos x ) we have2 y  = (1 − sin x ) − 3 / 2 cos x  = [(1 − sin x ) − 1 / 2 ] 3 cos x  =  y 3 cos x. An interval of definition for the solution of the differential equation is ( π/ 2 , 5 π/ 2). Another one is (5 π/ 2 , 9 π/ 2),and so on. 19.  Writing ln(2 X  − 1) − ln( X  − 1) =  t  and differentiating implicitly we obtain22 X  − 1 dX dt  −  1 X  − 1 dX dt  = 1   22 X  − 1  −  1 X  − 1   dX dt  = 12 X  − 2 − 2 X   + 1(2 X  − 1)( X  − 1) dX dt  = 1 dX dt  = − (2 X  − 1)( X  − 1) = ( X  − 1)(1 − 2 X  ) . Exponentiating both sides of the implicit solution we obtain2 X  − 1 X  − 1 =  e t 2 X  − 1 =  Xe t − e t ( e t − 1) = ( e t − 2) X X   =  e t − 1 e t − 2  . Solving  e t − 2 = 0 we get  t  = ln2. Thus, the solution is defined on ( −∞ , ln2) or on (ln2 , ∞ ). The graph of thesolution defined on ( −∞ , ln2) is dashed, and the graph of the solution defined on (ln2 , ∞ ) is solid. 2 www.librosysolucionarios.net
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