One-dimensional first-order inhomogenous linear ODEs are ODEs of the form
for suitable (that is, mostly, continuous) functions ; note that when , we have a homogenous equation instead.
First we note that we have the following superposition principle: If we have a solution ("" standing for "homogenous") of the problem
(which is nothing but the homogenous problem associated to the above ODE) and a solution to the actual problem ; that is a function such that
("" standing for "particular solution", indicating that this is only one of the many possible solutions), then the function
- ( arbitrary)
still solves , just like the particular solution does. This is proved by computing the derivative of directly.
In order to obtain the solutions to the ODE under consideration, we first solve the related homogenous problem; that is, first we look for such that
It may seem surprising, but this gives actually a very quick path to the general solution, which goes as follows. Separation of variables (and using ) gives
since the function
is an antiderivative of . Thus we have found the solution to the related homogenous problem.
For the determination of a solution to the actual equation, we now use an Ansatz: Namely we assume
where is a function. This Ansatz is called variation of the constant and is due to Leonhard Euler. If this equation holds for , let's see what condition on we get for to be a solution. We want
- , that is (by the product rule and inserting ):
Putting the exponential on the other side, that is
Since all the manipulations we did are reversible, all functions of the form
- ( arbitrary)
are solutions. If we set , we get the general solution form
We want now to prove that these constitute all the solutions to the equation under consideration. Thus, set
and let be any other solution to the inhomogenous problem under consideration. Then solves the homogenous problem, for
Thus, if we prove that all the homogenous solutions (and in particular the difference ) are of the form
then we may subtract
from for an appropriate to obtain zero, which is why is then of the desired form.
Thus, let be any solution to the homogenous problem. Consider the function
We differentiate this function and obtain by the product rule
since is a solution to the homogenous problem. Hence, the function is constant (that is, equal to a constant ), and solving
for gives the claim.
We have thus arrived at:
For continuous , the solutions to the ODE
are precisely the functions
- ( arbitrary).
Note that imposing an inital condition for some enforces , whence we got a unique solution for each initial condition.
- Exercise 3.2.1: First prove that . Then solve the ODE for a function existent on such that for arbitrary. Use that a similar version of theorem 3.1 holds when are only defined on a proper part of ; this is because the proof carries over.
Clever Ansatz for polynomial RHS
First note that RHS means "Right Hand Side". Let's consider the special case of a 1-dim. first-order linear ODE
- ( arbitrary),
where we used Einstein summation convention; that is, stands for for some . In the notation of above, we have and .
Using separation of variables, the solution to the corresponding homogenous problem is easily seen to equal for some capital .
To find a particular solution , we proceed as follows. We pick the Ansatz to assume that is simply a polynomial; that is
for certain coefficients .
- Exercise 3.3.1: Find all solutions to the ODE . (Hint: What does theorem 3.1 say about the number of solutions to that problem with a given fixed initial condition?)