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Problem using variation of parameters

  1. Apr 3, 2005 #1
    the problem is fin the general solution of the differential eq :

    y''+y=2sect + 3 (-pi/2 < t < pi/2)

    using variation of parameters.

    I just needed a check to make sure my answer was correct.

    r^2+1 = 0
    r= -i
    r= i
    y1= cost
    y2= sint
    g(t)= 2sect+ 3

    y(t) = c1cost + c2sint + Y(t)
    Y(t) = u1y1 + u2y2
    u1 = -(integal) (y2*g(t))/W in which W = 1
    = -(int) sint(2sect+ 3)
    = -(int) sint(2/cost+3)
    = -(2 (int) tant + 3 (int) sint)
    is this correct, where do i go from here

    u2= (integral) y1*g(t)/ W
    = (int) cost(2sect + 3)/ W
    = (int) (2*(cost/cost) + 3cost)
    = (int) 2 + (int)3 cost
    = 2+3(sint)
    is this correct

    and then I plug these back into the Y(t) eq and add this to y(t)?
     
  2. jcsd
  3. Apr 3, 2005 #2
    Yup. And your answer is correct.
     
  4. Apr 4, 2005 #3

    dextercioby

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    Science Advisor
    Homework Helper

    I dunno what u did,but here's where you have to get...

    [tex] \frac{d^2y}{dx^2}+y=2\sec x+3 [/tex]

    Exact solution is :

    [tex]y\left( x\right) =2\ln \left( 1-\cos x-\sin x\right) \cos x+3\cos x+2\ln \left( 1-\cos x+\sin x\right) \cos x-4\ln \left( \sin x\right) \cos x[/tex]
    [tex]\allowbreak +2\ln \left( \cos x+1\right) \cos x-2\ln 2\cos x+4\sin x\arctan \left(\frac{\sin x}{\cos x+1}\right)+3+C_1\cos x+\allowbreak C_2\sin x [/tex]

    Daniel.
     
    Last edited: Apr 4, 2005
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