# 1st order, nonhomogeneous, linear DE - particular solution

I know nothing about DEs, so this may be a silly question.

I'm given some time varying $(x_t)_t$ and a constant $r$, and I want to solve the equation $u_t = rx_t + \dot u_t$ for $u$.

What I know so far is that (solving the homogeneous equation) if $\bar u$ is some particular solution, then any $u$ is a solution iff it takes the form $u_t=\bar u_t + \alpha e^{rt}$ for some constant $\alpha$.

I'm wondering whether there's a brute force way of finding some $\bar u$. Anywhere I've looked suggests the "method of undetermined coefficients", but I know it's not useful in my setting. Is there a formula I can blindly apply to get a particular solution? I'm happy to assume $x$ as well behaved as needed, and I'm happy to have my formula be some horrible definite integral I can't compute.

## Answers and Replies

fzero
Science Advisor
Homework Helper
Gold Member
The general method for finding particular solutions is called variation of parameters, though in this case, we can use an integrating factor to save quite a bit of work. Note that

$$e^t \frac{d}{dt} \left( e^{-t} u\right) = \dot{u} - u,$$

so that the integrating factor ##e^{-t}## will lead to a particular solution in the form of an indefinite integral involving ##x_t##.

HallsofIvy
Science Advisor
Homework Helper
I know nothing about DEs, so this may be a silly question.

I'm given some time varying $(x_t)_t$ and a constant $r$, and I want to solve the equation $u_t = rx_t + \dot u_t$ for $u$.
So the subscript t just means that u and x are functions of t?

What I know so far is that (solving the homogeneous equation) if $\bar u$ is some particular solution, then any $u$ is a solution iff it takes the form $u_t=\bar u_t + \alpha e^{rt}$ for some constant $\alpha$.
No. That would be the case if the "r" were multiplying $u_t$, not x. The general solution to the equation $u'(t)+ rx(t)= u(t)$ is $\bar u(t)+ \alpha e^t$.

I'm wondering whether there's a brute force way of finding some $\bar u$. Anywhere I've looked suggests the "method of undetermined coefficients", but I know it's not useful in my setting. Is there a formula I can blindly apply to get a particular solution? I'm happy to assume $x$ as well behaved as needed, and I'm happy to have my formula be some horrible definite integral I can't compute.
I'm not sure you would call it "brute force" but I would use "variation of parameters". Knowing that $u(t)= Ce^t$, for C any constant, satisfies $u'- u= 0$, to solve the problem $u'- u= -rx(t)$, look for a solution of the form $u(t)= v(t)e^t$. Then $u'(t)= v'(t)e^t+ ve^t$. so $u'- u= v'e^t+ ve^t- ve^t= v'e^t= -rx$ then $v'= -re^{-t}x(t)$. Then $v(t)= -r\int_0^t e^{-\tau}x(\tau)d\tau+ C$ so that
$$u(t)= -re^t\int_0^t e^{-\tau}x(\tau)d\tau+ Ce^t$$

Whoops, I'd meant to write $ru_t = rx_t + \dot u_t$, but anyway...

Thank you very much, folks! You have been extremely helpful. :)