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Math Help - Basic Laplace Transform of a 1st order Differential Equation

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    Super Member craig's Avatar
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    Basic Laplace Transform of a 1st order Differential Equation

    Just wondering if someone could have a quick look over what I've done so far:

    Assuming zero initial conditions, solve the following DE:

    \dot{x} + x =  2

    The Laplace Transform gives:

    sX(s) - x_0 + \frac{1}{s^2} = \frac{2}{s}

    Rearranging and using the fact that x_0  =0 gives us

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

    X(s) = 2\frac{1!}{s^{1+1}} - \frac{1}{2} \frac{2!}{s^{2+1}}

    So x(s) = 2t - \frac{t^2}{2}?
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    After taking the Laplace transform, where did the \frac{1}{s^2} come from?
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    Super Member craig's Avatar
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by Danny View Post
    After taking the Laplace transform, where did the \frac{1}{s^2} come from?
    I took the Laplace transform of x, is this not right?
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by craig View Post
    I took the Laplace transform of x, is this not right?
    x is a dependent variable ( x=x(t) ), so you'll obtain

    s\mathcal{L}\{x\}-x_0+\mathcal{L}\{x\}=\frac{2}{s}\Rightarrow (s+1)\mathcal{L}\{x\}=x_0+\frac{2}{s}\Rightarrow \ldots
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by FernandoRevilla View Post
    x is a dependent variable ( x=x(t) ), so you'll obtain

    s\mathcal{L}\{x\}-x_0+\mathcal{L}\{x\}=\frac{2}{s}\Rightarrow (s+1)\mathcal{L}\{x\}=x_0+\frac{2}{s}\Rightarrow \ldots
    Thanks for the reply!

    So given that x_0 = 0, does that mean that we just have \mathcal{L}\{x\}=\frac{2}{s(s+1)}.

    Which I presume we express in partial fractions, and then solve for x(t)?
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    Super Member craig's Avatar
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    So for \mathcal{L}\{x\}=\frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1} we have:

    \mathcal{L}\{x\}= 2\frac{1}{s} - 2\frac{1}{s-(-1)}

    So x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t) ?

    Thanks again for the replies.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by craig View Post
    So x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t) ?
    Right.
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by craig View Post
    So for \mathcal{L}\{x\}=\frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1} we have:

    \mathcal{L}\{x\}= 2\frac{1}{s} - 2\frac{1}{s-(-1)}

    So x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t) ?

    Thanks again for the replies.
    Technically, since you're working with the one-sided LT (I assume), you have unit step functions multiplying everything. That is,

    x(t)=2u(t)(1-e^{-t}),

    and the solution is valid for non-negative t's.
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by Ackbeet View Post
    and the solution is valid for non-negative t's.
    Also considering the theorem about the necessary form of the solutions for the equation x^{(n)}+a_{n-1}x^{(n-1)}+\ldots +a_1x'+a_0x=b(t) .
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    Super Member craig's Avatar
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by FernandoRevilla View Post
    Also considering the theorem about the necessary form of the solutions for the equation x^{(n)}+a_{n-1}x^{(n-1)}+\ldots +a_1x'+a_0x=b(t) .
    Sorry would you mind explaining what you meant by this? I understood the first part about the Unit-Step function, just not sure what you meant here?

    Thankyou again!
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by craig View Post
    Sorry would you mind explaining what you meant by this?
    The Laplace transform of x:[0,+\infty)\to\mathbb{R} is defined by \mathcal{L}\{x(t)\}=\int_0^{+\infty}e^{-st}x(t)\;dt , so a priori we find solutions valid for t\geq 0. Your equation has (by a well known theorem) a determined form for its unique solution x(t)=k+Ce^{at} valid for all t\in\mathbb{R} so, necessarily is valid the solution found by LT for all t\in\mathbb{R} .
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Quote Originally Posted by FernandoRevilla View Post
    The Laplace transform of x:[0,+\infty)\to\mathbb{R} is defined by \mathcal{L}\{x(t)\}=\int_0^{+\infty}e^{-st}x(t)\;dt , so a priori we find solutions valid for t\geq 0. Your equation has (by a well known theorem) a determined form for its unique solution x(t)=k+Ce^{at} valid for all t\in\mathbb{R} so, necessarily is valid the solution found by LT for all t\in\mathbb{R} .
    Agreed. You could simply say that you extend the solution found by the LT method to the entire real line. The result of the inverse LT does have the unit step functions in it, and, in fact, does not satisfy the DE for negative t's. However, with the theorem you have invoked, you can extend the solution by eliminating the unit step function.
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    Super Member craig's Avatar
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    Re: Basic Laplace Transform of a 1st order Differential Equation

    Thankyou both! I don't think we've covered this in lectures yet but I think it makes sense.
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