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Math Help - Integrals and Equations Involving Step Functions

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    Integrals and Equations Involving Step Functions

    Hello Everyone!

    I've recently came across step functions in my systems course, they frequently appear in ODEs describing the system.

    An ODE I came across was y'+y=u(x). Now, I know I must ask this in the differential equations forum, but, I know what the solution should be, it's just that I'm not "okay" with integrals involving step functions. In the above equation, we would get y = \displaystyle \frac{\int e^x . u(x)dx}{e^x} but what's \displaystyle \int e^x .u(x)dx? Shouldn't it be plain u(x) . dx?
    If so, when substituting our new solution in the DE, we get: \delta (x) + u(x)  \neq u(x). How come?!

    Any help is appreciated!
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    Quote Originally Posted by rebghb View Post
    Hello Everyone!

    I've recently came across step functions in my systems course, they frequently appear in ODEs describing the system.

    An ODE I came across was y'+y=u(x). Now, I know I must ask this in the differential equations forum, but, I know what the solution should be, it's just that I'm not "okay" with integrals involving step functions. In the above equation, we would get y = \displaystyle \frac{\int e^x . u(x)dx}{e^x} but what's \displaystyle \int e^x .u(x)dx? Shouldn't it be plain u(x) . dx?
    If so, when substituting our new solution in the DE, we get: \delta (x)  + u(x)  + u(x)  \neq u(x). How come?!

    Any help is appreciated!
    Maybe the Maths people here have a way to handle it all at once, but my suggestion is to split the equation:
    1) On ( -\infty , 0) use y' + y = 0

    2) On ( 0, \infty ) use y' + y = 1.

    -Dan

    Reflecting upon this a little bit...You could use a Laplace transform to handle this in one equation.
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  3. #3
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    Quote Originally Posted by rebghb View Post
    Hello Everyone!

    I've recently came across step functions in my systems course, they frequently appear in ODEs describing the system.

    An ODE I came across was y'+y=u(x). Now, I know I must ask this in the differential equations forum, but, I know what the solution should be, it's just that I'm not "okay" with integrals involving step functions. In the above equation, we would get y = \displaystyle \frac{\int e^x . u(x)dx}{e^x} but what's \displaystyle \int e^x .u(x)dx? Shouldn't it be plain u(x) . dx?
    If so, when substituting our new solution in the DE, we get: \delta (x) + u(x) \neq u(x). How come?!

    Any help is appreciated!
    It's likely that posts here will duplicate the effort of posters here: http://www.mathhelpforum.com/math-he...de-172697.html

    Thread closed.

    (In future try not to post questions that have this much similarity).
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