hw_2_19_20_09

hw_2_19_20_09 - MA 221 Homework Solutions Due date:...

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MA 221 Homework Solutions Due date : February 19 - 20 , 2009 Section 4 . 6pg197 - 98 : 1 , 3 ̄ , 1 ̄ 1 ̄ , 1 ̄ 8 ̄ , 2 ̄ 1 ̄ Section 8 . 5 pg 460 1 ̄ , 3 , 5 ̄ , 1 ̄ 5 ̄ (Underlined problems are to be handed in) Section 4.6 pg 197-98 1.) Find a general solution to the differential equation using variation of parameters. y ′′ 4 y tan2 t This equation has associated homogenous equation y ′′ 4 y 0 The roots of the associated homogenous equation, r 2 4 0,a r e r  2 i . Therefore, a general solution to this equation is y h t c 1 cos2 t c 2 sin2 t For the variation of parameters method, we let y p t v 1 t y 1 t v 2 t y 2 t where y 1 t t and y 2 t t Thus, y 1 t 2sin2 t and y 2 t 2cos2 t . This means that we have to solve the system v 1 t v 2 t 0 2 v 1 t t 2 v 2 t t t Multiplying the first equation by s t the second equation by 1 2 t and adding the resulting equations together, we get: v 2 t 1 2 t v 2 1 2 tdt 1 4 t c 3 Substituting v 2 into the first equation v 1 t v 2 t 0 ,weget v 1 t v 2 t t 1 2 sin 2 2 t t 1 2 1 cos 2 2 t cos t 1 2 t sec2 t v 1 t 1 2 t t dt 1 4 t ln|sec2 t t | c 4 We take c 3 c 4 0 since we just need a particular solution. Thus, y p t 1 4 t ln|sec2 t t | t 1 4 t t 1 4 t ln|sec2 t t | and general solution is y p t y h t y p t c 1 t c 2 t 1 4 t ln|sec2 t t | 3 ̄ .) Find a general solution to the differential equation using variation of parameters. 2 x ′′ 2 x 4 x 2 e 3 t First, we can simplify the equation by dividing both sides by 2. This yields x ′′ x 2 x e 3 t This equation has associated homogenous equation x ′′ x 2 x 0 The roots of the associated homogenous equation, r 2 r 2 r e r 2an d r 1. Therefore, a general solution to this equation is 1
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x h t c 1 e 2 t c 2 e t
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hw_2_19_20_09 - MA 221 Homework Solutions Due date:...

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