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Math Help - System of DE's: Laplace transfomation

  1. #1
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    System of DE's: Laplace transfomation

    Using the Laplace transform, find the functions x(t) and y(t) that satisfy

    x=y=0 at t=0 and

    dx/dt + 2(dy/dt) + x = t
    -dx/dt - dy/dt + y =1
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  2. #2
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    Quote Originally Posted by Kbotz View Post
    Using the Laplace transform, find the functions x(t) and y(t) that satisfy

    x=y=0 at t=0 and

    dx/dt + 2(dy/dt) + x = t
    -dx/dt - dy/dt + y =1
    What have you tried? Where do you get stuck?
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  3. #3
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    i transformed each one of the variables.
    so for the 1st equation i got

    sX+2sY = 1/s^2 ......................... eq 3

    and transformation of the second equation
    -sX-sY = 1/s.............................. eq 4

    then eliminated X from both equations by multiplying (eq 3) by -1, and taking eq 4 from it.
    Then made Y the subject

    Y = 1+s/s^3

    I want to know if the previous steps are right, and the approach to the problem.

    Thanks Fantastic.
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  4. #4
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    Quote Originally Posted by Kbotz View Post
    i transformed each one of the variables.
    so for the 1st equation i got

    sX+2sY = 1/s^2 ......................... eq 3

    and transformation of the second equation
    -sX-sY = 1/s.............................. eq 4

    [snip]
    What about the Laplace transform of x and y in the original DE's?

    The correct equations are:

    sX + 2sY + X = 1/s^2 ......................... eq 3

    -sX - sY + Y = 1/s.............................. eq 4
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  5. #5
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    oh right, thanks
    i will work on it from here and post back if i need more help!
    thanks again for the heads up mr.fantastic!
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  6. #6
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    hey,
    i ran into more trouble doing this problem.
    i have no idea whether my answer is right.

    transformed equations as in earlier post
    <br />
sX + 2sY + X = 1/s^2 eqn (1)

    -sX - sY + Y= 1/s eqn (2)


    Multiply equation 1 by s^2 to remove the s^2 on the RHS.

    <br />
    s^3X + 2s^3Y + s^2X = 1 eqn (1')


    Multiply equation 2 by s to remove the 's' on the RHS.
    <br />
    -s^2x - s^2Y + sY = 1 eqn (2')

    equation 1' & 2' commonly factorized in terms of X

    <br />
X(s^3+s^2) + 2s^3Y = 1 .... eqn(1')
    X(-s^2) - s^2Y + sY = 1 .....eqn(2')

    To eliminate X from the pair of equations; eqn 2' is multiplied by (-s-1)

    After further simplification

    X(s^3 + s^2) + s^3Y - sY = -s-1.....eqn(2'')

    Now X can be eliminated, so subtract eqn (2'') from eqn (1')

    I don't know if this is right
    , after subtraction i got:

    s^3Y + sY = 2-s<br />
Y(s^3+s) = 2-s

    Make Y the subject

    Y = 2-s/(s^3+s), which is the same as  Y = 2-s/(s*(s^2+1))

    Now using partial fractions

    2-s = A/s + B/(s^2+1)

    cross multiply

    2-s = A(s^2+1) + B(s)

    when s = 0, B=0 and A = 2

    then when s =1, substitute in A =2 to find out B

    s = 1, B = -3

    Taking the inverse laplace transform of the partial fraction:

    Y = 2/s - 3/(s^2+1)

    y(t) = 2 - 3sin(t)


    Please check my steps, i hope it makes sense.
    And i have no idea how to get x(t), i keep coming to really complex equations.
    Last edited by Kbotz; May 2nd 2010 at 09:47 PM.
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  7. #7
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    Quote Originally Posted by Kbotz View Post
    hey,
    i ran into more trouble doing this problem.
    i have no idea whether my answer is right.

    transformed equations as in earlier post
    <br />
sX + 2sY + X = 1/s^2 eqn (1)

    -sX - sY + Y= 1/s eqn (2)


    Multiply equation 1 by s^2 to remove the s^2 on the RHS.

    <br />
s^3X + 2s^3Y + s^2X = 1 eqn (1')


    Multiply equation 2 by s to remove the 's' on the RHS.
    <br />
-s^2x - s^2Y + sY = 1 eqn (2')

    equation 1' & 2' commonly factorized in terms of X
    <br />
X(s^3+s^2) + 2s^3Y = 1 .... eqn(1')
    X(-s^2) - s^2Y + sY = 1 .....eqn(2')

    To eliminate X from the pair of equations; eqn 2' is multiplied by (-s-1)

    After further simplification

    X(s^3 + s^2) + s^3Y - sY = -s-1.....eqn(2'')

    Now X can be eliminated, so subtract eqn (2'') from eqn (1')

    I don't know if this is right, after subtraction i got:

    s^3Y + sY = 2-s<br />
Y(s^3+s) = 2-s

    Make Y the subject

    Y = 2-s/(s^3+s), which is the same as  Y = 2-s/(s*(s^2+1))

    Now using partial fractions

    2-s = A/s + B/(s^2+1)

    cross multiply

    2-s = A(s^2+1) + B(s)

    when s = 0, B=0 and A = 2

    then when s =1, substitute in A =2 to find out B

    s = 1, B = -3

    Taking the inverse laplace transform of the partial fraction:

    Y = 2/s - 3/(s^2+1)

    y(t) = 2 - 3sin(t)


    Please check my steps, i hope it makes sense.
    And i have no idea how to get x(t), i keep coming to really complex equations.
    I get:

    x = - \frac{(2s^2 + s - 1)}{s^2 (s^2 + 1)} = \frac{s}{s^2 + 1} - \frac{3}{s^2 + 1} - \frac{1}{s} + \frac{1}{s^2}


    y = \frac{s+2}{s (s^2 + 1)} = \frac{-2s}{s^2 + 1} + \frac{1}{s^2 + 1} + \frac{2}{s}.


    You just have to use basic algebra and solve the equations simultaneously for x and y and then express each solution in partial fraction form.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    I get:

    x = - \frac{(2s^2 + s - 1)}{s^2 (s^2 + 1)} = \frac{s}{s^2 + 1} - \frac{3}{s^2 + 1} - \frac{1}{s} + \frac{1}{s^2}


    y = \frac{s+2}{s (s^2 + 1)} = \frac{-2s}{s^2 + 1} + \frac{1}{s^2 + 1} + \frac{2}{s}.


    You just have to use basic algebra and solve the equations simultaneously for x and y and then express each solution in partial fraction form.
    i'm a little unsure of how you got s+2 for the y equation, but thanks for your help.
    i might try working backwards from your expression.
    cheers
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  9. #9
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    oh right, my bad!
    i can't subtract properly
    0-(-s) = s
    not -s. lol
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  10. #10
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    Quote Originally Posted by Kbotz View Post
    i'm a little unsure of how you got s+2 for the y equation, but thanks for your help.
    i might try working backwards from your expression.
    cheers
    Done correctly, s \times (1) + (s + 1) \times (2) will give you y.
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  11. #11
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    yeah figured them both out!
    and got the final transforms too!
    i was on the right track for x(t) from the start!
    but cheers again!
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