# Thread: Integrals and Equations Involving Step Functions

1. ## 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 $\displaystyle y'+y=u(x)$. 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 $\displaystyle y = \displaystyle \frac{\int e^x . u(x)dx}{e^x}$ but what's $\displaystyle \displaystyle \int e^x .u(x)dx$? Shouldn't it be plain $\displaystyle u(x) . dx$?
If so, when substituting our new solution in the DE, we get: $\displaystyle \delta (x) + u(x) \neq u(x)$. How come?!

Any help is appreciated!

2. 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 $\displaystyle y'+y=u(x)$. 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 $\displaystyle y = \displaystyle \frac{\int e^x . u(x)dx}{e^x}$ but what's $\displaystyle \displaystyle \int e^x .u(x)dx$? Shouldn't it be plain $\displaystyle u(x) . dx$?
If so, when substituting our new solution in the DE, we get: $\displaystyle \delta (x) + u(x) + u(x) \neq u(x)$. 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 $\displaystyle ( -\infty , 0)$ use y' + y = 0

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

-Dan

Reflecting upon this a little bit...You could use a Laplace transform to handle this in one equation.

3. 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 $\displaystyle y'+y=u(x)$. 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 $\displaystyle y = \displaystyle \frac{\int e^x . u(x)dx}{e^x}$ but what's $\displaystyle \displaystyle \int e^x .u(x)dx$? Shouldn't it be plain $\displaystyle u(x) . dx$?
If so, when substituting our new solution in the DE, we get: $\displaystyle \delta (x) + u(x) \neq u(x)$. 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