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Cal II basic question

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data
    y=sinxcosx-cos^{2}x Differential equation= 2y+y'=2sin(2x)-1
    Initial condition y(pi/4)=0






    3. The attempt at a solution

    2y=2sinxcosx-2cos^{2}x

    y'=-sin^{2}x+cos^{2}x+2cosxsinx

    ok... using double angle formulas... I get
    2y=Sin2x-2cos^{2}x

    y'=2cos^{2}x-1+2sinx

    2y+y'
    (Sin2x-2Cos^{2}x)+(2cos^{2}x-1+2sinx)
    the 2cos mess cancells out leaves you with

    Sin2x + (-1) +2Sinx
    I dont understand how that equals 2sin2x-1
    as the book says it does and the solution shows.

    Stupid question I know... I struggle with the easy stuff.

    Thanks
    -Ed
     
  2. jcsd
  3. Jun 22, 2008 #2
    You're trying to prove that [tex]sin(2x) + 2sin(x)-1 = 2sin(2x)-1 [/tex]

    Edit: I have to leave in a hurry, but this equality is not true.
    In the RHS, [tex]2sin(2x)-1=sin(2x)+sin(2x)-1[/tex]

    [tex]\implies sin(2x) + 2sin(x)-1 = sin(2x)+sin(2x)-1 [/tex]
    Therefore, cancelling the sin(2x)-1's from the LHS and RHS,
    [tex]2sin(x) \neq sin(2x)[/tex] since by the double angle formula,

    [tex]sin(2x) = 2sin(x)cos(x)[/tex]

    So, you probably made a mistake somewhere before.
     
    Last edited: Jun 22, 2008
  4. Jun 22, 2008 #3
    awesome... I'm totally screwed for tomorrow. lol

    Cal 2 during the summer was a terrible idea.
     
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