The Wronskian satisfies W y 1 y 2 t cos2 t sin2 t 2 sin2 t 2 cos2 t 2cos 2 2 t

# The wronskian satisfies w y 1 y 2 t cos2 t sin2 t 2

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The Wronskian satisfies: W [ y 1 , y 2 ]( t ) = cos(2 t ) sin(2 t ) - 2 sin(2 t ) 2 cos(2 t ) = 2(cos 2 (2 t ) + sin 2 (2 t )) = 2 . The variation of parameters gives y p ( t ) = - cos(2 t ) Z t 8 sin(2 s ) csc(2 s ) 2 ds + sin(2 t ) Z t 8 cos(2 s ) csc(2 s ) 2 ds = - 4 cos(2 t ) Z t 1 · ds + 4 sin(2 t ) Z t cos(2 s ) sin(2 s ) ds = - 4 t cos(2 t ) + 2 sin(2 t ) ln | sin(2 t ) | The general solution satisfies: y ( t ) = c 1 cos(2 t ) + c 2 sin(2 t ) - 4 t cos(2 t ) + 2 sin(2 t ) ln | sin(2 t ) | . e. The differential equation, y 00 + 4 y 0 + 4 y = e - 2 t t 2 , requires the variation of parameters. The characteristic equation is λ 2 + 4 λ + 4 = 0, so λ = - 2 (repeated root). Thus, the homogeneous solution is y c ( t ) = ( c 1 + c 2 t ) e - 2 t .

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The Wronskian satisfies: W [ y 1 , y 2 ]( t ) = e - 2 t te - 2 t - 2 e - 2 t (1 - 2 t ) e - 2 t = e - 4 t . The variation of parameters gives y p ( t ) = - e - 2 t Z t se - 2 s e - 2 s s 2 e - 4 s ds + te - 2 t Z t e - 2 s e - 2 s s 2 e - 4 s ds = - e - 2 t Z t 1 s ds + te - 2 t Z t 1 s 2 ds = - e - 2 t ln | t | - e - 2 t The general solution satisfies: y ( t ) = ( c 1 + c 2 t ) e - 2 t - e - 2 t (ln | t | + 1) . f. The differential equation, t 2 y 00 + 7 ty 0 + 5 y = 6 t , requires the variation of parameters. The left hand side is a Cauchy-Euler equation, so attempt solutions of the form y ( t ) = t λ , leaving a characteristic equation λ ( λ - 1) + 7 λ + 5 = λ 2 + 6 λ + 5 = 0 or λ 1 = - 5 and λ 2 = - 1. Thus, the homogeneous solution is y c ( t ) = c 1 t - 5 + c 2 t - 1 . The Wronskian satisfies: W [ y 1 , y 2 ]( t ) = t - 5 t - 1 - 5 t - 6 - t - 2 = 4 t - 7 . Note that the nonhomogeneous term is written g ( t ) = 6 t . The variation of parameters gives y p ( t ) = - t - 5 Z t 6( s - 1 ) 4 s ( s - 7 ) ds + t - 1 Z t 6( s - 5 ) 4 s ( s - 7 ) ds = - 3 2 t - 5 Z t s 5 ds + 3 2 t - 1 Z t sds = - 3 2 t 5 t 6 6 + 3 2 t t 2 2 = 1 2 t The general solution satisfies: y ( t ) = c 1 t - 5 + c 2 t - 1 + 1 2 t. 2. The differential equation is: (1 - t ) y 00 + ty 0 - y = 2( t - 1) 2 e - t or y 00 + t 1 - t y 0 - 1 1 - t y = - 2( t - 1) e - t . Since y 1 ( t ) = t , then y 0 1 ( t ) = 1 and y 00 1 ( t ) = 0. Substituting into the RHS of the equation above gives: 0 + t 1 - t - 1 1 - t t = 0 ,
so satisfies the homogeneous equation. Similarly, if y 2 ( t ) = e t , then y 0 2 ( t ) = e t and y 00 2 ( t ) = e t . Substituting into the RHS of the equation above gives: e t 1 - t 1 - t + te t 1 - t - e t 1 - t = 0 , so y 2 ( t ) satisfies the homogeneous problem. Solving the general problem requires finding a particular solution using the variation of parameters. The Wronskian satisfies: W [ y 1 , y 2 ]( t ) = t e t 1 e t = ( t - 1) e t . The variation of parameters gives y p ( t ) = - t Z t - e s (2( s - 1) e - s ) ( s - 1) e s ds + e t Z t - s (2( s - 1) e - s ) ( s - 1) e s ds = t Z t 2 e - s ds - e t Z t 2 se - 2 s ds = - 2 te - t + t + 1 2 e - t = 1 2 - t e - t The general solution satisfies: y ( t ) = c 1 t + c 2 e t + 1 2 - t e - t .

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