Integrals and Equations Involving Step Functions

**rebghb** · February 26th 2011, 10:06 AM

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!

**topsquark** · February 26th 2011, 10:11 AM

Originally Posted by rebghb

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.

**mr fantastic** · February 26th 2011, 01:11 PM

Originally Posted by rebghb

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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