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Thread: 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). $\displaystyle y''-2y'+y=7e^t cos(t)$

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

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

    So the equation I get is:
    $\displaystyle yparticular(t)=t*A*t^0*e^t*cos(0*t)+t*B*t^0*e^t*sin(0*t)$
    which simplifies to
    $\displaystyle y(t)=A*t*e^t$
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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). $\displaystyle y''-2y'+y=7e^t cos(t)$

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

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

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

    The plug it into the equation:

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

    where m=degree of t in original equation.
    s=0 if $\displaystyle \alpha + i \beta$ is not a root
    s=1 if $\displaystyle \alpha + i \beta$ 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). $\displaystyle y''-2y'+y=7e^t cos(t)$

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

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

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

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

    The plug it into the equation:

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

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

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

    $\displaystyle y_c = c_1 e^t + c_2 t e^t$

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

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

    Substitute gives

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

    Comparing gives A = -7 and B = 0 so

    $\displaystyle y_p = -7 e^t \cos t $

    and the solution

    $\displaystyle 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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