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Math Help - Laplace Transforms to solve First Order ODE's.

  1. #1
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    Laplace Transforms to solve First Order ODE's.

    Hey guys, first post. Looking forward to not only getting help here but also hopefully helping others in their quest for maths knowledge. I'm trying to do the following question but I am having all sorts of troubles!

    Solve the system of first order differential equations using Laplace Transforms:

    dx/dt = x - 4y

    dy/dt = x + y,
    subject to the initial conditions x(0)=3 and y(0)=-4.



    So far I've used the limited knowledge of Laplace Transforms for first order ODE's to get this far:

    L[x`] = L[x] - L[4y]
    s*L[x] - x(0) = L[x] - L[4y]
    s*L[x] - L[x] - 3 = -L[4y]
    (s-1)*L[x] = 3 + L[4y] <--------- Equation 1

    L[y'] = L[x] + L[y]
    s*L[y] - y(0) = L[x] + L[y]
    s*L[y] + 4 = L[x] + L[y]
    (s-1)*L[y] = L[x] - 4 <----------Equation 2
    or (s-1)*L[y] + 4 = L[x]


    Up to this stage I am kind of confident I have been using Laplace Transforms right (from the couple of examples I have in a text book I got from the library).




    The step where I become very confused is substituting equations 1 and 2 into one another to evaluate y(t) and x(t).

    When I substitute (2) into (1) i get the following:
    (s-1)[(s-1)*L[y] + 4] = 3 + L[4y]

    From here I have probably tried 20 different ways of getting a solution for y(s) but every single one is very complicated and leads to a dead end for me (they are way too long to type). Because of this I suspect I am doing something wrong here [potentially I am even applying Laplace Transforms completely wrong!].

    I am hoping someone knows where I am going wrong or what I'm doing wrong. Any help and advice would be greatly appreciated, thanks!
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  2. #2
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    Quote Originally Posted by WrittenStars View Post
    Hey guys, first post. Looking forward to not only getting help here but also hopefully helping others in their quest for maths knowledge. I'm trying to do the following question but I am having all sorts of troubles!

    Solve the system of first order differential equations using Laplace Transforms:

    dx/dt = x - 4y

    dy/dt = x + y,
    subject to the initial conditions x(0)=3 and y(0)=-4.



    So far I've used the limited knowledge of Laplace Transforms for first order ODE's to get this far:

    L[x`] = L[x] - L[4y]
    s*L[x] - x(0) = L[x] - L[4y]
    s*L[x] - L[x] - 3 = -L[4y]
    (s-1)*L[x] = 3 + L[4y] <--------- Equation 1

    L[y'] = L[x] + L[y]
    s*L[y] - y(0) = L[x] + L[y]
    s*L[y] + 4 = L[x] + L[y]
    (s-1)*L[y] = L[x] - 4 <----------Equation 2
    or (s-1)*L[y] + 4 = L[x]


    Up to this stage I am kind of confident I have been using Laplace Transforms right (from the couple of examples I have in a text book I got from the library).




    The step where I become very confused is substituting equations 1 and 2 into one another to evaluate y(t) and x(t).

    When I substitute (2) into (1) i get the following:
    (s-1)[(s-1)*L[y] + 4] = 3 + L[4y]

    From here I have probably tried 20 different ways of getting a solution for y(s) but every single one is very complicated and leads to a dead end for me (they are way too long to type). Because of this I suspect I am doing something wrong here [potentially I am even applying Laplace Transforms completely wrong!].

    I am hoping someone knows where I am going wrong or what I'm doing wrong. Any help and advice would be greatly appreciated, thanks!
    (s-1)[(s-1)*L[y] + 4] = 3 + L[4y]. But L[4y] = 4 L[y]. So (s-1)[(s-1)*L[y] + 4] = 3 + 4L[y].

    Solve for L[y] in terms of s and then take the inverse Laplace transform to get y. etc.
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  3. #3
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    Thanks so much for your reply! Really appreciate it.

    I isolated L[y] from your help to get:
    [3-4(s-1)] / [(s-1)^2 - 4]

    From here I am stuck again! I can't seem to find a way to isolate y and get a general solution for it. Sorry I'm not great with maths yet, just starting and keen.
    Last edited by WrittenStars; April 24th 2010 at 12:36 PM.
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  4. #4
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    Quote Originally Posted by WrittenStars View Post
    Thanks so much for your reply! Really appreciate it.

    I isolated L[y] from your help to get:
    [3-4(s-1)] / [(s-1)^2 - 4]

    From here I am stuck again! I can't seem to find a way to isolate y and get a general solution for it. Sorry I'm not great with maths yet, just starting and keen.
    First of all, simplify the numerator. Now break the expression up into tow parts. Now compare each part with the table of Laplace transforms you undoubtedly have.

    (If you're struggling with this question you should go back and practice on easier ones first).
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  5. #5
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    I simplified the numerator and got the following:

    (7-4s) / [(s-1)^2 - 4]

    From here I cant seem to find any way of separating the expression into 2 parts to resemble expressions in the Laplace transforms table.

    My textbook has some very simple questions that I've done successful but this one is just a extra 'harden question' that caught my eye. Sorry, I know I'm a little out of my league but I've spend quite some time on this question now and I just want to see how it works! Thanks for the help!
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  6. #6
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    Quote Originally Posted by WrittenStars View Post
    I simplified the numerator and got the following:

    (7-4s) / [(s-1)^2 - 4]

    From here I cant seem to find any way of separating the expression into 2 parts to resemble expressions in the Laplace transforms table.

    My textbook has some very simple questions that I've done successful but this one is just a extra 'harden question' that caught my eye. Sorry, I know I'm a little out of my league but I've spend quite some time on this question now and I just want to see how it works! Thanks for the help!
    Try partial fractions.

    \frac{7 - 4s}{(s - 1)^2 - 4} = \frac{7 - 4s}{(s - 1 + 2)(s - 1 - 2)}

     = \frac{7 - 4s}{(s + 1)(s - 3)}.


    So you have

    \frac{7 - 4s}{(s + 3)(s - 5)} = \frac{A}{s + 1} + \frac{B}{s - 3}

    \frac{7 - 4s}{(s + 3)(s - 5)} = \frac{A(s - 3) + B(s + 1)}{(s +1)(s - 3)}.


    Therefore

    A(s - 3) + B(s + 1) = 7 - 4s

    As - 3A + Bs + B = -4s + 7

    (A + B)s + B - 3A = -4s + 7.


    Therefore A + B = -4 and B - 3A = 7.

    From the first equation, A = -B - 4.

    So B - 3(-B - 4) = 7

    B + 3B + 12 = 7

    4B = -5

    B = -\frac{5}{4}.


    So A = \frac{5}{4} - 4 = -\frac{11}{4}.


    This means that

    \frac{7 - 4s}{(s + 1)(s - 3)} = -\frac{11}{4(s + 1)} - \frac{5}{4(s - 3)} .
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  7. #7
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    I went a different route and rather than simplifying the numerator, I started by applying the frequency shift rule:

    (L^-1)[(3-4(s-1)) / ((s-1)^2 - 4)] = (e^-t)(L^-1)[(3-4s)/ (s^2 - 4)]

    From there I have tried to now split the expression into 2 parts to use the Laplace Transformation table, by doing:

    (e^-t)*(L^-1)[3/(s^2 - 2^2)*(L^-1)[-4s/(s^2 - 2^2)]
    (e^-t)*[3/2](L^-1)[2/(s^2 - 2^2)*-4(L^-1)[s/(s^2 - 2^2)]

    From here I used the two following Laplace transformations:
    a/(s^2 + a^2) = sin(at) and s/(s^2 + a^2) = cos(at) to get:

    (e^-t)*[3/2]sin(-2t)*-4cos(-2t)
    -6(e^-t)*sin(-2t)*cos(-2t)

    Not sure if this is right, but it's the best application for the Laplace transformation table for this question I could see.

    I'll try the partial fractions method you have used now, thanks!
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  8. #8
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    Ok well going from getting:

    [-11/4(s+1)] - [5/4(s-3)]

    I figured I could use the Laplace transformation:
    1/(s+a) = exp(-at)

    so:
    (-11/4)*[1/(s+1)] - (5/4)*[1/(s-3)] = (-11/4)exp(-t) - (5/4)exp(3t)

    This seems like a better approach then what I posted previously!
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