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Math Help - inital value-problem

  1. #1
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    inital value-problem

    how would you solve this initial-value problem?

    \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{sin(4x)}{3+cos(4x)} when y=5 and x=0
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  2. #2
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    Re: inital value-problem

    Integrate both sides with respect to x. Then use the initial value pair to decide the value of the constant of integration.
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  3. #3
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    Re: inital value-problem

    Quote Originally Posted by Orlando View Post
    how would you solve this initial-value problem?

    \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{sin(4x)}{3+cos(4x)} when y=5 and x=0
    \int dy=\int \frac{\sin(4x)}{3+\cos(4x)} \,dx

    to solve this integral use substitution :

     \cos (4x) = u \Rightarrow du = - 4\sin(4x) dx
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  4. #4
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    Re: inital value-problem

    so

    y=\int\frac{-4sin(x4)}{3+cos(4x)}dx

    y=-\frac{1}{4}\int\frac{-4sin(4x)}{3+cos(4x)}dx

    so
    y=-\frac{1}{4}ln(3+cos(4x))+c

    so

    5=-\frac{1}{4}ln(3)+c

    so

    c=5+\frac{1}{4}ln(3)

    Im I on the right track?
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  5. #5
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    Re: inital value-problem

    Since ~\cos(0)=1~ fourth line should be :

    5= \frac{-1}{4} \cdot \ln(4) + C
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  6. #6
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    Re: inital value-problem

    sorry silly mistake

    so y=5-\frac{1}{4}ln(\frac{3+cos(4x)}{4})
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