1st order, nonhomogeneous, linear DE - particular solution

[economicsnerd]

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.

 

[fzero]

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]

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
$$u(t)= -re^t\int_0^t e^{-\tau}x(\tau)d\tau+ Ce^t$$

 

[economicsnerd]

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. :)

 

[blue_raver22]

I can provide some guidance on how to approach this problem. First, let's break down the terms in the equation. The "1st order" refers to the highest derivative in the equation, which in this case is u_t. "Nonhomogeneous" means that there is a term that is not equal to zero on the right side of the equation, in this case rx_t. Finally, "linear" means that the equation can be expressed as a sum of terms, each involving only a single variable and its derivatives.

To find a particular solution, one approach is to use the method of undetermined coefficients. However, as you mentioned, this may not be useful in your specific setting. Another approach is to use the method of variation of parameters, which can be applied to nonhomogeneous linear equations. This method involves finding a particular solution by assuming it has the form u_t = \bar u_t + \alpha(t), where \bar u is a particular solution and \alpha(t) is a function to be determined.

To find \bar u, you can use the method of undetermined coefficients to solve the homogeneous equation u_t = \dot u_t, which will give you a general solution in the form of u_t = \alpha e^{rt}. From there, you can use the method of variation of parameters to find a particular solution \bar u.

It is important to note that finding a particular solution can be a challenging task, and there may not always be a simple formula that can be blindly applied. It may require a combination of methods and techniques, and in some cases, the solution may involve integrals that cannot be easily computed. As a scientist, it is important to approach these problems with patience and persistence, and to use the appropriate tools and methods to find a solution.