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Math Help - Undetermined coefficients

  1. #1
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    Undetermined coefficients

    I am just wanting somebody to check my work.

    5. Find the Particular solution of the following equation using Undetermined Coefficients(yparticular(t). y''-2y'+y=7e^t cos(t)

    The auxiliary equation I get is:
    r^2-2r+1=0
    (r-1)^2=0
    r=1 Twice

    So,
    \alpha = \frac2 2 = 1
    \beta = \sqrt(4-4)/2) = 0
    That means that s=1

    So the equation I get is:
    particular(t)=t*A*t^0*e^t*cos(0*t)+t*B*t^0*e^t*sin(0*t)" alt="yparticular(t)=t*A*t^0*e^t*cos(0*t)+t*B*t^0*e^t*sin(0*t)" />
    which simplifies to
     y(t)=A*t*e^t
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  2. #2
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    Quote Originally Posted by jazz836062 View Post
    I am just wanting somebody to check my work.

    5. Find the Particular solution of the following equation using Undetermined Coefficients(yparticular(t). y''-2y'+y=7e^t cos(t)

    The auxiliary equation I get is:
    r^2-2r+1=0
    (r-1)^2=0
    r=1 Twice

    So,
    \alpha = \frac2 2 = 1
    \beta = \sqrt(4-4)/2) = 0
    And \alpha \;\text{and}\; \beta are?
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    And \alpha \;\text{and}\; \beta are?
    In my book the method of undetermined Coefficients calls for  \alpha and \beta.

    \alpha = \frac{-b} {2a}
    \beta = \frac{\sqrt{4ac-b^2}} {2a}

    The plug it into the equation:

    y_{particular}(t)=t^s(A_{m}t^m+...+A_{0})e^{\alpha t}cos(\beta t)+t^s(B_{m}t^m+...+B_{0})e^{\alpha t}sin(\beta t)

    where m=degree of t in original equation.
    s=0 if \alpha + i \beta is not a root
    s=1 if \alpha + i \beta is a root
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  4. #4
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    Quote Originally Posted by jazz836062 View Post
    I am just wanting somebody to check my work.

    5. Find the Particular solution of the following equation using Undetermined Coefficients(yparticular(t). y''-2y'+y=7e^t cos(t)

    The auxiliary equation I get is:
    r^2-2r+1=0
    (r-1)^2=0
    r=1 Twice

    So,
    \alpha = \frac2 2 = 1
    \beta = \sqrt(4-4)/2) = 0
    That means that s=1

    So the equation I get is:
    y_p(t)=t*A*t^0*e^t*cos(0*t)+t*B*t^0*e^t*sin(0*t)
    which simplifies to
     y(t)=A*t*e^t
    Quote Originally Posted by jazz836062 View Post
    In my book the method of undetermined Coefficients calls for \alpha and \beta.

    \alpha = \frac{-b} {2a}
    \beta = \frac{\sqrt{4ac-b^2}} {2a}

    The plug it into the equation:

    y_{particular}(t)=t^s(A_{m}t^m+...+A_{0})e^{\alpha t}cos(\beta t)+t^s(B_{m}t^m+...+B_{0})e^{\alpha t}sin(\beta t)

    where m=degree of t in original equation.
    s=0 if \alpha + i \beta is not a root
    s=1 if \alpha + i \beta is a root
    Instead of memorizing formula's, you could do the following.

    First, the complementary y''-2y'+y=0. As you said, the roots are 1, 1 so

    y_c = c_1 e^t + c_2 t e^t

    For the particular, ask is the nonhomogenous part contained in the complementary solution - no. So seek a particular solution of the form

    y_p = A e^t \cos t + B e^t \sin t

    Substitute gives

    -A e^t \cos t - B e^t \sin t = 7 e^t \cos t

    Comparing gives A = -7 and B = 0 so

    y_p = -7 e^t \cos t

    and the solution

    y = y_c + y_p = c_1 e^t + c_2 t e^t -7 e^t \cos t
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  5. #5
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    Thank you. I got it wrong on the test. I was going back and checking my answers after the fact and i got it way wrong. Thank you for your help.
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