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maths

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give the solution of the initial-value problem
dy/dx=(1+2cos^2(x))/y, (y > 0), y= 1 when x = 0.
Thanks.

  • maths -

    Is it an answer on my question?

    2cos^2(x)=1+cos(2x)
    ydy=(2+cos(2x))dx integrating
    y^2/2=2x+sin(2x)/2+C Find C
    1/2=C

    y=sqrt(4x+sin(2x)+1)

  • maths -

    Thats right.
    Thank you.

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