hw_2_17_09

# hw_2_17_09 - MA 221 Homework Solutions Due date Section 4.5...

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MA 221 Homework Solutions Due date : February 17 , 2009 Section 4 . 5pg . 192 2 ̄ b ̄ , 5 , 7 ̄ , 2 ̄ 1 ̄ , 2 ̄ 7 ̄ , 2 ̄ 9 ̄ (Underlined Problems are to be handed in) 2 ̄ .) Given that y 1 t 1/4 sin2 t is a solution to y ′′ 2 y 4 y cos2 t and that y 2 t t /4 1/8 is a solution to y ′′ 2 y 4 y t , use the superposition principle to find soltuions of (b) y ′′ 2 y 4 y 2 t 3cos2 t Since the right hand side of this last equation is twice the right hand side of the second equation given plus 3 times the right hand side of the first equation given a particular solution for (b) is 2 y 2 3 y 1 t 2 1 4 3 4 sin2 t general solution for the equation. 5.) ′′ 2 1 2 t p t t 1 The homogeneous equation is ′′ 2 0 Thus r 2 r 2 r 2  r 1 0so r 2, 1 y h c 1 e 2 t c 2 e t Therefore by the superposition principle: y h p t 1 c 1 e 2 t c 2 e t SNB check: ′′ 2 1 2 t , (Compute – Solve ODE – Exact – Independent Variable t ) Exact solution is: t C 3 e t C 4 e 2 t 1 7 ̄ .) y ′′ 2 y y 2 e x y p x x 2 e x The homogeneous equation is y ′′ 2 y y 0 Thus r 2 2 r 1 r 1 2 0so r 1 is a repeated root and y h c 1 e x c 2 xe x Therefore y y h y p c 1 e x c 2 xe x x 2 e x SNB check: y ′′ 2 y y 2 e x (Compute – Solve ODE – Exact – Independent Variable x ) Exact solution is: C 1 e x C 2 xe x x 2 e x In problem 21, find a general solution to the differential equation. 2

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hw_2_17_09 - MA 221 Homework Solutions Due date Section 4.5...

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