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Math Help - Undetermined Coefficients

  1. #1
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    Undetermined Coefficients

    I have a problem solving the undetermined coefficient problems. I know how to get the particular solution. But the algebra always messes me up.

    The problem is:
    y'' + 9y = t^2*e^3t + 6
    The particular solution y particular = (At^2 + Bt + C)e^3t + D.

    The next one is:
    2y'' + 3y' + y = t^2 + 3sint
    The particular solution y particular = (At^2 + Bt + C) + Asint + Bcost

    This problem is an initial value problem:
    y'' + 4y = 3sint2t, y(0) = 2, y'(0) = -1
    The particular solution y particular was the same as the homogeneous solution so I added a factor t to the particular solution. So my y particular was: t(Asin2t + Bcos2t).

    Tell me if these particular solutions are correct. My main problem was to solve for the constants like A, B, and even C. Can you show the step of how I can solve this. Even though I collected the terms but I get a strange equation that equals to the g(t).
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Here's the second one
    Quote Originally Posted by fudawala View Post
    The next one is:
    2y'' + 3y' + y = t^2 + 3sint
    The particular solution y particular = (At^2 + Bt + C) + Asint + Bcost
    2y'' + 3y' + y = t^2 + 3 \sin t

    For the homogeneous solution, try y = e^{ \lambda t}, thus we get the characteristic equation:

    2 \lambda ^2 + 3 \lambda + 1 = 0

    \Rightarrow \lambda = - \frac {1}{2} \mbox { or } \lambda = -1

    So, y_h = Ae^{-t} + Be^{-t/2}

    For a particular solution, try y_p = At^2 + Bt + C + D \sin t + E \cos t ........I hope you notice your mistake here. You reused A and B for the sine and cosine functions, you can't do that, they should be different variables!

    \Rightarrow y_p ' = 2At + B + D \cos t - E \sin t

    \Rightarrow y_p'' = 2A - D \sin t - E \cos t

    Plug these into the original equation, we get:

    4A - 2D \sin t - 2E \cos t + 6At + 3B + 3D \cos t - 3E \sin t + At^2  + Bt + C + D \sin t + E \cos t = t^2 + 3 \sin t

    Collect like terms:

    \Rightarrow (4A + 3B + C) + (6A + B)t + At^2 + (-3E - D) \sin t + (3D - E) \cos t = t^2 + 3 \sin t

    Equating coefficients we get:

    \boxed { A = 1}

    \Rightarrow 3B + C = -4

    \boxed { B = - 6} \implies \boxed { C = 14 }

    -D - 3E = 3

    3D - E = 0 \implies E = 3D

    \Rightarrow -D - 9D = 3 \implies \boxed { D = - \frac {3}{10} } \implies \boxed { E = - \frac {9}{10}}

    Thus the general solution is:

    y_g = y_h + y_p

    \Rightarrow y_g = Ae^{-t} + Be^{-t/2} + t^2 - 6t + 14 - \frac {3}{10} \sin t - \frac {9}{10} \cos t
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Here's the first

    Quote Originally Posted by fudawala View Post

    The problem is:
    y'' + 9y = t^2*e^3t + 6
    The particular solution y particular = (At^2 + Bt + C)e^3t + D.
    Good job getting the particular solution for that, now let's see how it all works out.

    y'' + 9y = t^2 e^{3t} + 6

    For the homogeneous solution, try y = e^{ \lambda t}, thus we get the characteristic equation:

    \lambda ^2 + 9 = 0

    \Rightarrow \lambda = \pm 3~i

    \Rightarrow y_h = A \cos 3t + B \sin 3t

    For the particular solution, try y_p = \left( At^2 + Bt + C \right)e^{3t} + D

    \Rightarrow y_p' = (2At + B) e^{3t} + 3 \left( At^2 + Bt + C \right) e^{3t}

    \Rightarrow y_p'' = 2Ae^{3t} + 3(2At + B)e^{3t} + 3(2At + B)e^{3t} + 9 \left( At^2 + Bt + C \right)e^{3t}

    Plug these into the original equation (you may simplify them first if you wish, but I'll jump right into it, I will also expand the brackets at the same time, remember, we must multiply y_p by 9), we get:

    2Ae^{3t} + 6Ate^{3t} + 3Be^{3t} + 6Ate^{3t} + 3Be^{3t} + 9At^2 e^{3t} + 9Bte^{3t} + 9C e^{3t} + 9At^2 e^{3t} + 9Bte^{3t} + 9Ce^{3t} + 9D = t^2 e^{3t} + 6

    Collect like terms:

    \Rightarrow (2A + 6B + 18C)e^{3t} + (12A + 18B)te^{3t} + 18At^2 e^{3t} + 9D = t^2 e^{3t} + 6

    Equating coefficients, we get:

    2A + 6B + 18C = 0

    12A + 18B = 0

    18A = 1 \implies \boxed { A = \frac {1}{18} }  \implies \boxed { B = - \frac {1}{27} }

    plug A and B into the first equation, we get \boxed { C = \frac {1}{162} }

    9D = 6 \implies \boxed { D = \frac {2}{3} }

    So our general solution is:

    y_g = y_h + y_p

    \Rightarrow y_g = A \cos 3t + B \sin 3t + \left( \frac {1}{18}  t^2 - \frac {1}{27} t + \frac {1}{162} \right)e^{3t} + \frac {2}{3}

    Or

    y_g = A \cos 3t + B \sin 3t + \frac {1}{162} \left( 9t^2 - 6t + 1 \right)e^{3t} + \frac {2}{3}
    Last edited by Jhevon; July 4th 2007 at 08:18 PM.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fudawala View Post
    This problem is an initial value problem:
    y'' + 4y = 3sint2t, y(0) = 2, y'(0) = -1
    The particular solution y particular was the same as the homogeneous solution so I added a factor t to the particular solution. So my y particular was: t(Asin2t + Bcos2t).
    Okay, let's see how far we get with this one.

    y'' + 4y = 3 \sin 2t y(0) = 2 \mbox { , } y'(0) = -1

    For the homogeneous solution, try y = e^{ \lambda t}, thus we get the characteristic equation:

    \lambda ^2 + 4 = 0

    \Rightarrow \lambda = \pm 2i

    \Rightarrow y_h = A \cos 2t + B \sin 2t

    For the particular solution we guess y_p = At \sin 2t + Bt \cos 2t

    \Rightarrow y_p' = A \sin 2t + 2At \cos 2t + B \cos 2t = 2Bt \sin 2t

    \Rightarrow y_p'' = 2A \cos 2t + 2A \cos 2t - 4At \sin 2t - 2B \sin 2t - 2B \sin 2t - 4Bt \cos 2t

    Plugging these into the original equation, we get:

    2A \cos 2t + 2A \cos 2t - 4At \sin 2t - 2B \sin 2t - 2B \sin 2t - 4Bt \cos 2t + 4At \sin 2t + 4Bt \cos 2t = 3 \sin 2t

    This simplifies to:

    4A \cos 2t - 4B \sin 2t = 3 \sin 2t

    Equating the coefficients we get:

    4A = 0 \implies \boxed { A = 0 }

    -4B = 3 \implies \boxed { B = - \frac {3}{4} }

    Thus, our general solution is of the form:

    y_g = y_h + y_p

    \Rightarrow y_g = y(t) = A \cos 2t + B \sin 2t - \frac {3}{4} t \cos 2t

    \Rightarrow y'(t) = -2A \sin 2t + 2B \cos 2t - \frac {3}{4} \cos 2t + \frac {3}{2} \sin 2t

    we are given the initial data that y(0) = 2 \mbox { and } y'(0) = -1

    \Rightarrow y(0) = A = 2 \implies \boxed { A = 2}

    \Rightarrow y'(0) = 2B - \frac {3}{4} = -1 \implies \boxed {  B = - \frac {1}{8}}

    And so, finally, our general solution is:

    y(t) = 2 \cos 2t - \frac {1}{8} \sin 2t - \frac {3}{4} t \cos 2t

    Now that wasn't so bad, was it?
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  5. #5
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    Method of Undetermined Coefficients

    Hey Jhevon,

    The question that you solved:
    2y'' + 3y' + y = t^2 + 3 \sin t

    How did you equate the equation 3B + C = -4. I don't understand where did you get that from.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fudawala View Post
    Hey Jhevon,

    The question that you solved:
    2y'' + 3y' + y = t^2 + 3 \sin t

    How did you equate the equation 3B + C = -4. I don't understand where did you get that from.
    Notice that after we collected the terms and wanted to equate coefficients, we had 4A + 3B + C being equated to the constant term on the right side of the equation, however, there was no constant term on the right. thus we have:

    4A + 3B + C = 0

    i found that A = 1 from another term right of the bat, so we now have

    4 + 3B + C = 0

    subtracting 4 from both sides we obtain

    3B + C = -4

    which is the equation in question

    sorry i didn't show all the steps, these problems are too much typing to begin with, i got lazy
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