HW4solutions - Differential Equations Assignment...

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Differential Equations Assignment # 4(Solutions) Variation of Parameters and Mechanical Vibrations 1. Use Variation of Parameters to solve the following non-homogeneous initial value problems: (a) y - 3 y + 2 y = 2 e t , y (0) = 0, y (0) = 0 Two homogeneous solutions to this problem are given by: y 1 = e t , y 2 = e 2 t The Wronskian of these solutions is given by: W ( y 1 , y 2 ) = e t e 2 t e t 2 e 2 t = e 3 t The variation of parameters formula gives the solution to this problem as: y = - y 1 t 0 y 2 ( s ) F ( s ) W ( y 1 , y 2 ) ds + y 2 t 0 y 1 ( s ) F ( s ) W ( y 1 , y 2 ) ds y = - e t t 0 e 2 s (2 e s ) e 3 s ds + e 2 t t 0 e s ( s + 1) e 3 s ds y = - 2 e t t 0 1 ds + 2 e 2 t t 0 e - s ds y = - 2 te t + 2 e 2 t - e - t + 1 y = - 2 te t - 2 e t + 2 e 2 t (b) y - 2 y + y = t + 1, y (0) = 0, y (0) = 0 Two homogeneous solutions to this problem are given by: y 1 = e t , y 2 = te t The Wronskian of these solutions is given by: W ( y 1 , y 2 ) = e t te t e t e t + te t = e 2 t 1
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The variation of parameters formula gives the solution to this problem as: y = - y 1 t 0 y 2 ( s ) F ( s ) W ( y 1 , y 2 ) ds + y 2 t 0 y 1 ( s ) F ( s ) W ( y 1 , y 2 ) ds y = - e t t 0 se s ( s + 1) e 2 s ds + te t t 0 e s ( s + 1) e 2 s ds y = - e t t 0 ( s 2 + s ) e - s ds + te t t 0 ( s + 1) e - s ds Integration by parts on each of these integrands yields the solution as: y = - e t - t 2 e - t - 3 te - t - 3 e - t + 3 + te t - te - t - 2 e - t + 2 y = t 2 + 3 t + 3 - 3 e t + - t 2 - 2 t + 2 te t y = t + 3 - 3 e t + 2 te t (c) y - y = e - t , y (0) = 0, y (0) = 0 Two homogeneous solutions to this problem are given by: y 1 = e t , y 2 = e - t The Wronskian of these solutions is given by: W ( y 1 , y 2 ) = e t e - t e t - e - t = 2 The variation of parameters formula gives the solution to this problem as: y = - y 1 t 0 y 2 ( s ) F ( s ) W ( y 1 , y 2 ) ds + y 2 t 0 y 1 ( s ) F ( s ) W ( y 1 , y 2 ) ds y = - e t t 0 e - s ( e - t ) 2 ds + e - t t 0 e s ( e - s ) 2 ds y = - e t 2 t 0 e - 2 s ds + e - t 2 t 0 1 ds Integration on each of these integrands yields the solution as: y = - e t 2 - e - 2 t 2 + 1 2 + e - t 2 [ t - 0] y = e - t 4 - e t 4 + te - t 2 2
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(d) y + y = sin( t ), y ( π ) = 0, y ( π ) = 0 Two homogeneous solutions to this problem are given by: y 1 = cos( t ) , y 2 = sin( t ) The Wronskian of these solutions is given by: W ( y 1 , y 2 ) = cos( t ) sin( t ) - sin( t ) cos( t ) = 1 The variation of parameters formula gives the solution to this problem as: y = - y 1 t 0 y 2 ( s ) F ( s ) W ( y 1 , y 2 ) ds + y 2 t 0 y 1 ( s ) F ( s ) W ( y 1 , y 2 ) ds y = - cos( t ) t 0 sin( s ) sin( s ) ds + sin( t ) t 0 cos( s ) sin( s ) ds The first term requires a trigonometric identity, while the second integrand is amenable to a u-substitution. Setting u = sin( s ) , du = cos( s ) y = - cos( t ) t 0 1 - cos(2 s ) 2 ds + sin( t ) s s =0 = tudu y = - cos( t ) s 2 - sin(2 s ) 4 | t 0 + sin( t ) u 2 2 | s s =0 = t y = - cos( t ) t 2 - sin(2 t ) 4 + sin( t ) sin 2 ( t ) 2
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