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Math Help - Please help with this First Order Differential Equation

  1. #1
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    Please help with this First Order Differential Equation

    Im having some trouble to solve this differential equation. If somebody knows the solution of this please reply.

    So the equation is:

    y'+y*sinX=0

    I started to solve it but im stucked with a transformation problem. See below. If my method is incorrect please fix it
    Attached Thumbnails Attached Thumbnails Please help with this First Order Differential Equation-dsc0489512.jpg  
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  2. #2
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    Quote Originally Posted by gergofoto View Post
    Im having some trouble to solve this differential equation. If somebody knows the solution of this please reply.

    So the equation is:

    y'+y*sinX=0

    I started to solve it but im stucked with a transformation problem. See below. If my method is incorrect please fix it
    then \ln y = \cos x - C

    y = e^{\cos x - C}, let e^{-C} = A

    then

    y = Ae^{\cos x}
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  3. #3
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    Quote Originally Posted by gergofoto View Post
    Im having some trouble to solve this differential equation. If somebody knows the solution of this please reply.

    So the equation is:

    y'+y*sinX=0

    I started to solve it but im stucked with a transformation problem. See below. If my method is incorrect please fix it
    It looks good so far

    -\ln(y)=-\cos(x)+C

    First mult both sides by -1 to get

    \ln(y)=\cos(x)+C Since C is arbitary we don't need the negative sign.

    Since the inverse of the log function is the exponential we get

    e^{\ln(y)}=e^{\cos(x)+C} Now using exponential properties we get

    y=e^{\cos(x)}\cdot e^{C}

    Once again since C is arbitary we can rename e^{C}=A to get

    y=Ae^{cos(x)}
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