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    Using the method of variation of parameters, solve the equation
    y'' + y = tanx, 0<x<pi/2

    The auxiliary equation is r^2 + 1 = 0 with roots (+/-)i so the solution of y'' + y = 0 is (c1)sinx + (c2)cosx

    So we seek a solution of the form
    y = (u1)sinx + (u2)cosx, where (u1) and (u2) are arbitrary functions of x.
    y' = ((u1)'sinx + (u2)'cosx) + ((u1)cosx - (u2)sinx)

    [1] set (u1)'sinx + (u2)'cosx = 0

    y'' = (u1)'cosx - (u2)'sinx - (u1)sinx - (u2)cosx

    [2] y'' + y = (u1)'cosx - (u2)'sinx = tanx

    Now this is the part I don't get:
    Solving equations [1] and [2], we get
    (u1)'(sin^2(x) + cos^2(x)) = cosxtanx
    (u1)' = sinx, (u1) = -cosx

    What did they do with equations [1] and [2] to eliminate (u2)' ???
    And how did they get (u1)' from (u1)'(sin^2(x) + cos^2(x)) = cosxtanx
    How did they even get (u1)'(sin^2(x) + cos^2(x)) = cosxtanx in the first place?!:confused: ....

    I bet it's something really silly that I didn't see.......
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    (u1)' sin(x) + (u2)' cos(x) = 0
    => (u1)' sin^2(x) + (u2)' cos(x) sin(x) = 0 ... (*)
    (multiply by sin(x))

    (u1)' cos(x) - (u2)' sin(x) = tan(x)
    => (u1)' cos^2(x) - (u2)' cos(x) sin(x) = cos(x)tan(x) ... (**)
    (multiply by cos(x))

    Now add (*) and (**) to get:
    (u1)' sin^2(x) + (u1)' cos^2(x) = cos(x)tan(x)
    => (u1)' (sin^2(x) + cos^2(x)) = cos(x)tan(x)
    => (u1)' = cos(x) * sin(x)/cos(x) = sin(x)
    (since sin^2(x) + cos^2(x) = 1)
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    (Original post by dvs)
    (u1)' sin(x) + (u2)' cos(x) = 0
    => (u1)' sin^2(x) + (u2)' cos(x) sin(x) = 0 ... (*)
    (multiply by sin(x))

    (u1)' cos(x) - (u2)' sin(x) = tan(x)
    => (u1)' cos^2(x) - (u2)' cos(x) sin(x) = cos(x)tan(x) ... (**)
    (multiply by cos(x))

    Now add (*) and (**) to get:
    (u1)' sin^2(x) + (u1)' cos^2(x) = cos(x)tan(x)
    => (u1)' (sin^2(x) + cos^2(x)) = cos(x)tan(x)
    => (u1)' = cos(x) * sin(x)/cos(x) = sin(x)
    (since sin^2(x) + cos^2(x) = 1)
    You legend! For both helping me out and having a strange sleeping pattern like me , what time do you usually sleep?!
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    (Original post by [email protected])
    You legend! For both helping me out and having a strange sleeping pattern like me , what time do you usually sleep?!
    I go to sleep around 4-5am.

    Although to be fair, right now it's almost 11pm (different time zones and all that).
 
 
 
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Updated: April 13, 2006
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