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Math Help - Solution of Differential equations by Laplace Transforms

  1. #1
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    Solution of Differential equations by Laplace Transforms

    I am new to this area of maths and would like to know from the following equation:

    dx/dt -2x = 4 with conditions of t = 0 and x = 1

    when rewritten to give sx - 1 - 2x = 4/s how do you rearrange to give an expression for x?

    I have the answer as s + 4/s(s-2) but dont know how it has come to this.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by hunterage2000 View Post
    I am new to this area of maths and would like to know from the following equation:

    dx/dt -2x = 4 with conditions of t = 0 and x = 1

    when rewritten to give sx - 1 - 2x = 4/s how do you rearrange to give an expression for x?

    I have the answer as s + 4/s(s-2) but dont know how it has come to this.

    Thanks in advance
    Your notation is terrible. x represents the function in the time domain. You just cannot use the same symbol to represent the Laplace transform of x.

    And you do not have the answer as s + 4/s(s-2). Wherever you're getting the answer from will say \displaystyle \frac{4 + s}{s(s-2)}, which you should have written as (s+4)/s(s-2) if you're not going to typset your equations. s + 4/s(s-2) does not mean \displaystyle \frac{4 + s}{s(s-2)}, it means \displaystyle s + \frac{4}{s(s-2)}.

    Let LT(x) = X.

    Then, using basic algebra (which you're expected be competent in if you're studying at a level that includes Laplace transforms):

    \displaystyle sX - 1 - 2X = \frac{4}{s} \Rightarrow X(s - 2) - 1 = \frac{4}{s} \Rightarrow X (s - 2) = \frac{4}{s} + 1 = \frac{4 + s}{s} etc.
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  3. #3
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    In terms of Laplace Transform the DE becomes...

    \displaystyle s\ X(s) - x(0) - 2\ X(s) = \frac{4}{s} (1)

    ... so that [using pure algebraic rules as explained by Mr Fantastic...] is...

    \displaystyle X(s) = \frac{s+4}{s\ (s-2)} = \frac{3}{s-2} - \frac{2}{s} (2)

    ... and then...

    \displaystyle x(t)= \mathcal{L}^{-1} \{X(s)\} = 3\ e^{2 t} - 2 (3)

    Kind regards

    \chi \sigma
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    How about this equation

    sx - 6 + 2x = 10 / s-3 rearranged to:

    x = (6s -8) / (s + 2)(s - 3)

    Where has the -8 has come from?
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    Quote Originally Posted by hunterage2000 View Post
    sx - 6 + 2x = 10 / s-3 rearranged to:

    x = (6s -8) / (s + 2)(s - 3)

    Where has the -8 has come from?
    It arises from the same basic algebra as I used in post #2. I suggest you attempt this algebra yourself. If you need help, post all your work and say where you get stuck.
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    I am working through Advanced Engineering Mathematics by K.A. Stroud and have an example which is solve dx/dt + 2x = 10e^3t.

    I have sx -6 + 2x = 10 / s - 3 which I have factorized sx and 2x to give x(s + 2) - 6.

    I have taken (s + 2) then -6 over to give 6s / (s + 2)(s - 3). I'm just wondering where the -8 comes into it.
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  7. #7
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    Quote Originally Posted by hunterage2000 View Post
    I am working through Advanced Engineering Mathematics by K.A. Stroud and have an example which is solve dx/dt + 2x = 10e^3t.

    I have sx -6 + 2x = 10 / s - 3 which I have factorized sx and 2x to give x(s + 2) - 6.

    I have taken (s + 2) then -6 over to give 6s / (s + 2)(s - 3). I'm just wondering where the -8 comes into it.
    You have \displaystyle X (s + 2) - 6 = \frac{10}{s - 3}. The stuff I have highlighted in red bears no resemblance to the algebra I showed you in post #2. The red stuff is wrong. Please try and apply basic algebraic manipulations.
    Last edited by mr fantastic; August 14th 2010 at 05:35 AM.
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    I have taken everything over and still dont know where the -8 comes from. There is no 8 on the left hand side to take over.
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  9. #9
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    Quote Originally Posted by hunterage2000 View Post
    I have taken everything over and still dont know where the -8 comes from. There is no 8 on the left hand side to take over.
    I can only say this so many times. Take greater care with the algebra.

    Don't say what you did. Show it. Show the algebra. Show every single step. Show every single step of how you go from \displaystyle X (s + 2) - 6 = \frac{10}{s - 3} to making X the subject.

    Show every detail.
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  10. #10
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    Yeah I have

    1). x (s + 2) -6 = 10 / s - 3 - take over the (s + 2) to give

    2) x -6 = 10 / (s + 2)(s - 3) - take over the -6 to give

    3) x = 10 / (s + 2)(s - 3) + 6

    now not sure what to do
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  11. #11
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    Quote Originally Posted by hunterage2000 View Post
    Yeah I have

    1). x (s + 2) -6 = 10 / s - 3 - take over the (s + 2) to give Mr F says: This is not what I did in post #2. It is wrong. Are you familiar with order of operations in basic arithmetic? The correct thing to do is to first add 6 to both sides to get what I have posted below.

    2) x -6 = 10 / (s + 2)(s - 3) - take over the -6 to give

    3) x = 10 / (s + 2)(s - 3) + 6

    now not sure what to do
    \displaystyle X (s + 2) = \frac{10}{s - 3} + 6 = ....

    where .... means get a common denominator, simplify etc.

    If you hope to answer these sorts of questions correctly you will need to become competent with basic algebra.
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    Can you show me how you would rearrange it please?
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  13. #13
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    Quote Originally Posted by hunterage2000 View Post
    Can you show me how you would rearrange it please?
    I just don't understand how you can be studying at the level of Laplace transforms and not have a grasp of basic algebra.

    I have given you the next step in the algebra. And told you what to do next. And I gave you a solution to a very similar question in post #2.

    It's your turn again to show what comes next. Do you know what it means to get a common denominator? You are strongly advised to review basic algebra because otherwise you are in for a world of pain.
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