# Thread: Particular Solution for Constant Coefficients ODE

1. ## Particular Solution for Constant Coefficients ODE

Hello Everyone!

I'm trying to recall the basics of ODEs, I'm using that in my systems course. Suppose, we have an equation:
$\displaystyle y'+y=f(x)$
The solution would be 2 contributions, the solution to the homogenous equation and the particular solutions, now my questions are:

(1) Does the particular solution have to be of the form $\displaystyle f(x)$?
(2) Take $\displaystyle f(x)$ to be the unit step function $\displaystyle u(x)$, what would then the solution be?

Thanks for the help?

2. (1) Typically, you take your guess from the form of f(x).

(2) I would use the standard integrating factor method for solving this equation. The integrating factor is just $\displaystyle e^{x},$ so you'd get the following:

$\displaystyle [e^{x}y]'=e^{x}f(x)$

$\displaystyle \displaystyle e^{x}y=\int u(x)e^{x}\,dx=(e^{x}-1)\,u(x)+C\dots$

3. I tried substituting that but didn't work... Tried this

$\displaystyle y'e^x + ye^x = u(x)e^x$ becomes $\displaystyle (ye^x)'=u(x)e^x$
so $\displaystyle y=\displaystyle \frac{\int e^x . u(x) dx}{e^x}$ okay but what's $\displaystyle \displaystyle \int e^x . u(x) dx$?

4. Originally Posted by rebghb
I tried substituting that but didn't work... Tried this

$\displaystyle y'e^x + ye^x = u(x)e^x$ becomes $\displaystyle (ye^x)'=u(x)e^x$
so $\displaystyle y=\displaystyle \frac{\int e^x . u(x) dx}{e^x}$ okay but what's $\displaystyle \int e^x . u(x) dx$?
It's the far RHS of the second equation of my first post.