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Math Help - Initial Value Problem?

  1. #1
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    Initial Value Problem?

    dy/dx=(y-3)^2sin(3x) with the initial condition of f(1)=0. Find the particular solution. I need serious help!
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by xvaliant View Post
    dy/dx=(y-3)^2sin(3x) with the initial condition of f(1)=0. Find the particular solution. I need serious help!
    Separate the Variables:

    \frac{dy}{(y-3)^2} = sin(3x) \:dx

    Let u = y-3 so that du = dy

    \int \frac{du}{u^2} = \int sin(3x) \:dx

    When you solved that put in x=1 and y=0 to find a value for the constant of integration, C
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    Quote Originally Posted by e^(i*pi) View Post
    Separate the Variables:

    \frac{dy}{(y-3)^2} = sin(3x) \:dx

    Let u = y-3 so that du = dy

    \int \frac{du}{u^2} = \int sin(3x) \:dx

    When you solved that put in x=1 and y=0 to find a value for the constant of integration, C
    Okay, so I get (y-3)^2=C(1)e^((-1/3)cos3x) Then plug in x=1 and y=0 to get C. 9=C(1)(.7168) So C(1)=12.5558? So then I plug it back into the equation to get:

    (y-3)^2=12.5558e^((-1/3)cos3x)?
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    Quote Originally Posted by xvaliant View Post
    Okay, so I get (y-3)^2=C(1)e^((-1/3)cos3x) Then plug in x=1 and y=0 to get C. 9=C(1)(.7168) So C(1)=12.5558? So then I plug it back into the equation to get:

    (y-3)^2=12.5558e^((-1/3)cos3x)?
    Where did you get the exponential from?

    \int frac{du}{u^2} = -\frac{1}{u}

    As u = y-3: -\frac{1}{u} = -\frac{1}{y-3}

    The RHS is a standard trig integral, pull out the 1/3 to get it as -\frac{1}{3}cos(3x) + C

    Overall I get:

    -\frac{1}{y-3} = -\frac{1}{3}cos(3x) + C

    Since f(1)=0

    -\frac{1}{0-3} = \frac{1}{3} = -frac{1}{3}cos(3) + C

    C = \frac{1}{3}(1+cos(3))
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  5. #5
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    Oh! I get it now. Thanks so much!
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