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Can any one help with the following please:
Solve the initial-value problem:
problem:
dy/dx: 4(3cos(6x)-1)/1+2x-sin(6x) y=12 when x=0

  • Calc-IVP -

    Do you mean
    dy/dx= 4(3cos(6x)-1) / (1+2x-sin(6x)) ?

    Substitute
    u=(1+2x-sin(6x))
    du=(-6cos(6x)+2)

    so
    dy/dx = -2∫du/u
    y=-2ln(1+2x-sin(6x))+C

    Initial condition: y(0)=12
    y(0)=12=-2ln(1+2(0)-sin(6(0)))+C
    12=-2ln(1)+C
    12=0+C
    => C=12
    So
    y=-2ln(1+2x-sin(6x))+12

    Check the result by differentiation.

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