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Math Help - Differential Equations Application Problems

  1. #1
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    Differential Equations Application Problems

    Hi
    Need help on a few questions:

    1) The equation of motion of a mass (on the end of a vertical spring) subject to an "input force" (which causes a forced motion) is

    \frac{d^2y}{dt^2} + 9y = 5cos(3t)

    Find y given that when t=0, y=0 and y' = 0

    I get
    y = y_p +y_h

    y = \frac{5}{6}tsin(3t) - \frac{5}{6}sin(3t)

    book's answer is  y= \frac{5}{6}tsin(3t)

    2) a) A particle moving in a straight line has displacement x, velocity v and acceleration a at time t, where a=\frac{dv}{dt}=v\frac{dv}{dx}. These formulas can be used to find v given a. The former is used when v is required in terms of t, the latter for v in terms of x.

    b) A particle moves with retardation proportional to its velocity, ie, a = -kv, and has initial velocity v_0 and initial displacement 0. Find the velocity (i) in terms of time; and (ii) in terms of the displacement.

    How would i do this question?

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Need help on a few questions:

    1) The equation of motion of a mass (on the end of a vertical spring) subject to an "input force" (which causes a forced motion) is

    \frac{d^2y}{dt^2} + 9y = 5cos(3t)

    Find y given that when t=0, y=0 and y' = 0

    I get
    y = y_p +y_h

    y = \frac{5}{6}tsin(3t) - \frac{5}{6}sin(3t) Mr F says: Does this 'solution' satisfy y' = 0 ....? If you post your calculations, your mistake(s) can be pointed out.

    book's answer is  y= \frac{5}{6}tsin(3t)

    2) a) A particle moving in a straight line has displacement x, velocity v and acceleration a at time t, where a=\frac{dv}{dt}=v\frac{dv}{dx}. These formulas can be used to find v given a. The former is used when v is required in terms of t, the latter for v in terms of x.

    b) A particle moves with retardation proportional to its velocity, ie, a = -kv, and has initial velocity v_0 and initial displacement 0. Find the velocity (i) in terms of time; and (ii) in terms of the displacement.

    How would i do this question?

    P.S
    2) b) (i) Solve dv/dt = -kv subject to v(0) = v_0.

    2) b) (ii) Solve v dv/dx = -kv subject to v = v_0 when x = 0.
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  3. #3
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    ok, heres my workings:

    y_h = Acos(3t)+Bsin(3t)

    y_p = Atsin(3t)+Btcos(3t)

    y_p' = 3tcos(3t)+Asin(3t)-3Btsin(3t)+Bcos(3t)
    y_p'' = -9Atsin(3t)+3Acos(3t)+3Acos(3t)-9Btcos(3t)-3Bsin(3t)-3Bsin(3t)
    y_p'' = -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)

    sub back into original equation:

    -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)+ 9Atsin(3t)+9Btcos(3t) = 5cos(3t)
    6Acos(3t)-6Bsin(3t)=5cos(3t)=5cos(3t)
    6A=5
    A=\frac{5}{6}

    -6B=0
    B=0

    y_p = \frac{5}{6}tsin(3t)

    t=0, y=0, y'=0

    for y(0)=0
    0=Acos0+Bsin0+\frac{5}{6}sin0
    A=0

    y'(0)=0

    y'=-3Asin(3t)+3Bcos(3t)+\frac{15}{6}cos(3t)
    0=-3Asin(30)+3Bcos(30)+\frac{15}{6}cos(30)
    3B=\frac{-15}{6}
    B=\frac{-15}{6}*\frac{1}{3}

    y_h=\frac{-5}{6}sin(3t)

    y=\frac{5}{6}tsin(3t)-\frac{5}{6}sin(3t)
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    ok, heres my workings:

    y_h = Acos(3t)+Bsin(3t)

    y_p = Atsin(3t)+Btcos(3t)

    y_p' = 3tcos(3t)+Asin(3t)-3Btsin(3t)+Bcos(3t)
    y_p'' = -9Atsin(3t)+3Acos(3t)+3Acos(3t)-9Btcos(3t)-3Bsin(3t)-3Bsin(3t)
    y_p'' = -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)

    sub back into original equation:

    -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)+ 9Atsin(3t)+9Btcos(3t) = 5cos(3t)
    6Acos(3t)-6Bsin(3t)=5cos(3t)=5cos(3t)
    6A=5
    A=\frac{5}{6}

    -6B=0
    B=0

    y_p = \frac{5}{6}tsin(3t)

    t=0, y=0, y'=0

    for y(0)=0
    0=Acos0+Bsin0+\frac{5}{6}sin0
    A=0

    y'(0)=0

    y'=-3Asin(3t)+3Bcos(3t)+\frac{15}{6}cos(3t)
    0=-3Asin(30)+3Bcos(30)+\frac{15}{6}cos(30)
    3B=\frac{-15}{6}
    B=\frac{-15}{6}*\frac{1}{3}

    y_h=\frac{-5}{6}sin(3t)

    y=\frac{5}{6}tsin(3t)-\frac{5}{6}sin(3t)
    Your calculation is good and will benefit many members. Sorry that I misread your original post. The book is wrong.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    2) b) (i) Solve dv/dt = -kv subject to v(0) = v_0.
    this is what i got for i)
    \frac{dv}{dt}=-kv

    v=\frac{-kv^2}{2}+C
    v(0)=v_0
    v_0=C

    therefore i get \frac{-kv^2}{2}+v_0
    however this is not correct.
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  6. #6
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    Quote Originally Posted by Paymemoney View Post
    this is what i got for i)
    \frac{dv}{dt}=-kv

    v=\frac{-kv^2}{2}+C
    v(0)=v_0
    v_0=C

    therefore i get \frac{-kv^2}{2}+v_0
    however this is not correct.
    You have integrated the right hand side with respect to v and the left hand side with respect to t. You're expected to know the technique for this type of DE:

    The next step \frac{dt}{dv}=- \frac{1}{kv} and then you integrate both sides with respect to v ....
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