maths

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Looking for some guidance please, I need to find the initial-value problem for dy/dx = 1+2cos^2x / y. y = 1 when x = 0.
any help is much appricated

  • maths -

    the hard part is to integrate the cos^2x

    we know that cos 2x = 2cos^2x - 1
    so replace the 2cos^2x with cos2x + 1

    then
    dy/dx = 1+2cos^2x
    = 1 + cos2x + 1 = 2 + cos2x

    then y = 2x + (1/2)sin2x + c
    given: when x=0, y = 1
    1 = 0 + sin0 + c
    c = 1

    y = 2x + (1/2)sin2x + 1

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