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Math Help - Differential equation and its y(t) answer given a u(t) statement

  1. #1
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    Differential equation and its y(t) answer given a u(t) statement

    Given the differential equation:
    \frac{d^{4}y}{dt^{4}}+6\frac{d^{2}y}{dt^{2}}+25y=u  '+3u

    Find the y(t) solution when u(t)=e^{t}\cos t
    Hint: e^{t}\cos t=Re(e^{(1+i)t})

    The answer should be:
    y(t)=\frac{32}{195}e^{t}\cos t+\frac{9}{195}e^{t}\sin t

    My attempt (I'm not sure if I'm using the correct method):
    I start by using the "guessing method" which states that if you have u(t)=sin(kt) then you can guess on the solution with the form:
    y(t)=A \cos(kt)+B \sin(kt)

    So my guess so far is:
    y(t)=A \cos(t)+B \sin(t)

    I hereby differentiate this guess 2 and 4 times and insert it to my original equation:

    y''(t)=-A \cos(t)-B \sin(t)
    y''''(t)=A \cos(t)+B \sin(t)

    Inserted:
    (A \cos(t)+B \sin(t))+6\cdot (-A \cos(t)-B \sin(t))+25\cdot (A \cos(t)+B \sin(t))=u'+3u

    I clean it up a little:
    20\cdot A \cos(t)+20\cdot B \sin(t)=u'+3u

    This is as far as I've made it because in the examples in my textbook, they have not shown what to do when u appears in the original equation, thus I have no idea how to isolate and find the values of A and B.

    Can somebody please tell me if I'm on right track at all, and if so, what do I do next?
    Last edited by MathIsOhSoHard; September 5th 2012 at 11:33 AM.
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  2. #2
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    Re: Differential equation and its y(t) answer given a u(t) statement

    Hey MathIsSoHard.

    This is a linear ordinary differential equation (since you have the explicit form for u(t)).

    Try using Differential Operator techniques. Let P(D) = (D^4 + 6D^2 + 25). You will need to solve 1/P(D)[u' + 3u] for the specific solution p_s and solve P(D)y = 0 using the substitution y = e^(mx) for the complementary solution p_c.

    Then your solution will be the sum of these giving y(t) = p_c + p_s (which are both functions of t).
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    Re: Differential equation and its y(t) answer given a u(t) statement

    Quote Originally Posted by chiro View Post
    Hey MathIsSoHard.

    This is a linear ordinary differential equation (since you have the explicit form for u(t)).

    Try using Differential Operator techniques. Let P(D) = (D^4 + 6D^2 + 25). You will need to solve 1/P(D)[u' + 3u] for the specific solution p_s and solve P(D)y = 0 using the substitution y = e^(mx) for the complementary solution p_c.

    Then your solution will be the sum of these giving y(t) = p_c + p_s (which are both functions of t).
    I've worked a little more on it, this is how far I've made it so far:

    Given the differential equation:
    \frac{d^{4}y}{dt^{4}}+6\frac{d^{2}y}{dt^{2}}+25y=u  '+3u

    And we have that u(t)=e^{t}\cos(t)

    If I substitute u(t)=e^{t}\cos t on the right side of the equation, I get:
    \frac{d^{4}y}{dt^{4}}+6\frac{d^{2}y}{dt^{2}}+25y=e  ^t \cdot \cos(t) - e^t \cdot \sin(t) + 3 \cdot e^t \cdot \cos(t)

    I start by using the "guessing method" which states that if you have u(t)=sin(kt) then you can guess on the solution with the form:
    y(t)=A \cos(kt)+B \sin(kt)
    And if you have u(t)=e^{st}

    So my guess so far is:
    y(t)=A\cdot e^t \cdot \cos(t)+B \cdot e^t \cdot \sin(t)

    I hereby differentiate this guess 2 and 4 times and insert it to my original equation:

    y''(t)=-2 \cdot A \cdot e^t \cdot \sin(t) + 2 \cdot B \cdot e^t \cdot \cos(t)
    y''''(t)=-4 \cdot A \cdot e^t \cdot \cos(t) - 4 \cdot B \cdot e^t \cdot \sin(t)

    Inserted into the original equation:
    -4 \cdot A \cdot e^t \cdot \cos(t) - 4 \cdot B \cdot e^t \cdot \sin(t)+6\cdot (-2 \cdot A \cdot e^t \cdot \sin(t) + 2 \cdot B \cdot e^t \cdot \cos(t))=-e^t \cdot \sin(t) + 4 \cdot e^t \cdot \cos(t)

    I clean it up a little:
    (-4A+12B+25A)\cdot e^t \cdot \cos(t) + (-4B-12A+25B)\cdot e^t \sin(t)=-e^t \cdot \sin(t) + 4 \cdot e^t \cdot \cos(t)

    The coefficients of the trigonometric functions is:
    -4A+12B+25A=4
    -4B-12A+25B=-1

    Two equations with two unknown factors (A and B)
    and I get the values to:
    A=\frac{32}{195}
    B=\frac{3}{65}

    The answer should be:
    y(t)=\frac{32}{195}e^{t}\cos t+\frac{9}{195}e^{t}\sin t
    So my A value is correct, but my B value should be \frac{9}{195}???
    Can someone point out to me where my error is?
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    Re: Differential equation and its y(t) answer given a u(t) statement

    I feel like such an idiot.

    \frac{3}{65} is the same as \frac{9}{195}

    This is what happens when you lack severe sleep.... I guess I solved this question then...
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