hw5soln-corrected

# hw5soln-corrected - [Name[Section Homework#5 Math 527 1 The...

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[Name] [Section #] Homework #5 Math 527 1 The equation to solve is ° D 2 + 4 ± y = sin( x ). (1.1) Solve the associated homogeneous equation. ° D 2 + 4 ± y = 0 ( D 2 i )( D 2 i ) y = 0 y = C 1 cos(2 x ) + C 2 sin(2 x ) (1.2) Now make the judicious guess y = α cos( x ) + β x (1.3) Then ° D 2 + 4 ± y = ° α x ) β x ) ± + 4 ° α x ) + β x ) ± = 3 α x ) + 3 β x ); substituting this expression into (1.1) gives 3 α x ) + 3 β x ) = x ), which implies α = 0 and β = 1 3 . Substitute these values into (1.3). y = 1 3 x ) The sum of this solution with (1.2) is the general solution. y = 1 3 x ) + C 1 cos(2 x ) + C 2 sin(2 x ) 1

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2 The equation to solve is ° D 2 + 4 ± y = sin(2 x ). (2.1) We already have the homogeneous solution in (1.2). The judicious guess y = α cos(2 x ) + β sin(2 x ) won’t work, since it’s already a homoge- neous solution. So instead use y = x ° α cos(2 x ) + β sin(2 x ) ± . (2.2) Then, by the product rule, D 2 y = 2 ° 2 α sin(2 x ) + 2 β cos(2 x ) ± + ° 4 α cos(2 x ) 4 β sin(2 x ) ± = 4 β cos(2 x ) 4 α sin(2 x ) 4 x ° α cos(2 x ) + β cos(2 x ) ± , so ° D 2 + 4 ± y = 4 β cos(2 x ) 4 α sin(2 x Substituting this expression into (2.1) gives 4 β cos(2 x ) 4 α sin(2 x ) = sin(2 x Solve for α and β . 4 β = 0 and 4 α = 1 β = 0 and α =− 1 4 Substitute these values into (2.2). y 1 4 x cos(2 x ) The sum of this solution with (1.2) is the general solution. y 1 4 x cos(2 x ) + C 1 cos(2 x ) + C 2 sin(2 x ) 2
3 The equation to solve is ° D 2 4 D + 10 ± y = e t sin( t ). (3.1) Solve the associated homogeneous equation. ° D 2 4 D + 10 ± y = 0 Solve the characteristic equation. λ 2 4 λ + 10 = 0 λ = 4 ± ° 16 40 2 = 2 ± i ° 6 These are two nonreal roots. So, y = e 2 t ² C 1 cos ² ° 6 t ³ + C 2 sin ² ° 6 t ³³ . (3.2) Now, make the judicious guess y = e t ° α cos( t ) + β t ) ± . (3.3) For convenience, substitute s = α t ) + β t ). Then y = e t s ; Dy = e t ° s ² s ± ; D 2 y = e

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hw5soln-corrected - [Name[Section Homework#5 Math 527 1 The...

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