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:
$y'+y=f(x)$
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 $f(x)$?
(2) Take $f(x)$ to be the unit step function $u(x)$, 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 $e^{x},$ so you'd get the following:

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

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

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

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

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

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