# Ordinary Differential Equations/One-dimensional first-order linear equations
## Definition

One-dimensional first-order inhomogenous linear ODEs are ODEs of the form
    $x'(t)+f(t)x(t)=g(t)$
for suitable (that is, mostly, continuous) functions $f,g:\mathbb {R} \to \mathbb {R}$; note that when $g\equiv 0$, we have a homogenous equation instead.

## General solution

First we note that we have the following superposition principle: If we have a solution $x_{h}$ ("$h$" standing for "homogenous") of the problem
    $x_{h}'(t)+f(t)x_{h}(t)=0$
(which is nothing but the homogenous problem associated to the above ODE) and a solution to the actual problem $x_{p}$; that is a function $x_{p}$ such that
    $x_{p}'(t)+f(t)x_{p}(t)=g(t)$
("$p$" standing for "particular solution", indicating that this is only one of the many possible solutions), then the function
    $x(t):=ax_{h}(t)+x_{p}(t)$ ($a\in \mathbb {R}$ arbitrary)
still solves $x'(t)+f(t)x(t)=g(t)$, just like the particular solution $x_{p}$ does. This is proved by computing the derivative of $x$ 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 $x_{h}$ such that
    $x_{h}'(t)+f(t)x_{h}(t)=0\Leftrightarrow x_{h}'=-f(t)x_{h}$.
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 $\ln ^{-1}=\exp$) gives
    $x_{h}(t)=\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)$,
since the function
    $G(t):=-\int _{t_{0}}^{t}f(s)ds$
is an antiderivative of $t\mapsto -f(t)$. Thus we have found the solution to the related homogenous problem.
For the determination of a solution $x_{p}$ to the actual equation, we now use an Ansatz: Namely we assume
    $x_{p}(t)=c(t)x_{h}(t)$,
where $c:\mathbb {R} \to \mathbb {R}$ is a function. This Ansatz is called variation of the constant and is due to Leonhard Euler. If this equation holds for $x_{p}$, let's see what condition on $c$ we get for $x_{p}$ to be a solution. We want
    $x_{p}'(t)+f(t)x_{p}(t)=g(t)$, that is (by the product rule and inserting $x_{h}$):
    $c'(t)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)=c'(t)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)+c(t)(-1)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)f(t)+f(t)c(t)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)=g(t)$.
Putting the exponential on the other side, that is
    $c'(t)=g(t)\exp \left(\int _{t_{0}}^{t}f(s)ds\right)$
or
    $c(t)=\int _{t_{0}}^{t}g(r)\exp \left(\int _{t_{0}}^{r}f(s)ds\right)dr+C_{1}$.
Since all the manipulations we did are reversible, all functions of the form
    $C_{2}\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)+\left(\int _{t_{0}}^{t}g(r)\exp \left(\int _{t_{0}}^{r}f(s)ds\right)dr+C_{1}\right)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)$ ($C_{1},C_{2}\in \mathbb {R}$ arbitrary)
are solutions. If we set $C:=C_{2}+C_{1}$, we get the general solution form
    $C\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)+\left(\int _{t_{0}}^{t}g(r)\exp \left(\int _{t_{0}}^{r}f(s)ds\right)dr\right)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)$.
We want now to prove that these constitute all the solutions to the equation under consideration. Thus, set
    $x_{C}(t):=C\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)+\left(\int _{t_{0}}^{t}g(r)\exp \left(\int _{t_{0}}^{r}f(s)ds\right)dr\right)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)$
and let $x_{2}(t)$ be any other solution to the inhomogenous problem under consideration. Then $x_{C}-x_{2}$ solves the homogenous problem, for
    $x_{C}'(t)-x_{2}'(t)-f(t)(x_{C}(t)-x_{2}(t))=x_{C}'(t)-f(t)x_{C}(t)-(x_{2}'(t)-f(t)x_{2}(t))=g(t)-g(t)=0$.
Thus, if we prove that all the homogenous solutions (and in particular the difference $x_{C}-x_{2}$) are of the form
    $C\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)$,
then we may subtract
    $D\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)$
from $x_{C}-x_{2}$ for an appropriate $D\in \mathbb {R}$ to obtain zero, which is why $x_{2}$ is then of the desired form.
Thus, let $x_{h}$ be any solution to the homogenous problem. Consider the function
    $t\mapsto x_{h}(t)\cdot \exp \left(\int _{t_{0}}^{t}f(s)ds\right)$.
We differentiate this function and obtain by the product rule
    $x_{h}'(t)\exp \left(\int _{t_{0}}^{t}f(s)ds\right)+f(t)\exp \left(\int _{t_{0}}^{t}f(s)ds\right)x_{h}(t)=-f(t)x_{h}(t)\exp \left(\int _{t_{0}}^{t}f(s)ds\right)+f(t)\exp \left(\int _{t_{0}}^{t}f(s)ds\right)x_{h}(t)=0$
since $x_{h}$ is a solution to the homogenous problem. Hence, the function is constant (that is, equal to a constant $C\in \mathbb {R}$), and solving
    $x_{h}(t)\cdot \exp \left(\int _{t_{0}}^{t}f(s)ds\right)=C$
for $x_{h}$ gives the claim.
We have thus arrived at:
Theorem 3.1:
For continuous $f,g:\mathbb {R} \to \mathbb {R}$, the solutions to the ODE
    $x'(t)+f(t)x(t)=g(t)$
are precisely the functions
    $x(t)=C\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)+\left(\int _{t_{0}}^{t}g(r)\exp \left(\int _{t_{0}}^{r}f(s)ds\right)dr\right)\exp \left(-\int _{t_{0}}^{t}f(s)ds\right)$ ($C\in \mathbb {R}$ arbitrary).
Note that imposing a condition $x(t_{0})=x_{0}$ for some $x_{0}\in \mathbb {R}$ enforces $C=x_{0}$, whence we got a unique solution for each initial condition.

### Exercises

* Exercise 3.2.1: First prove that ${\frac {d}{dt}}\ln(t^{2})={\frac {2}{t}}$. Then solve the ODE $x'(t)+{\frac {2}{t}}x(t)={\frac {1}{t^{2}}}$ for a function existent on $[1,\infty )$ such that $x(1)=c$ for $c\in \mathbb {R}$ arbitrary. Use that a similar version of theorem 3.1 holds when $f,g$ are only defined on a proper part of $\mathbb {R}$; 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
    $x'(t)+cx(t)=a_{i}t^{i}$ ($c\in \mathbb {R}$ arbitrary),
where we used Einstein summation convention; that is, $a_{i}x^{i}$ stands for $\sum _{i=0}^{m}a_{i}t^{i}$ for some $m\in \mathbb {N}$. In the notation of above, we have $f(t)\equiv c$ and $g(t)=a_{i}t^{i}$.
Using separation of variables, the solution to the corresponding homogenous problem $g\equiv 0$ is easily seen to equal $x_{h}(t)=C\exp(-ct)$ for some capital $C\in \mathbb {R}$.
To find a particular solution $x_{p}$, we proceed as follows. We pick the Ansatz to assume that $x_{p}$ is simply a polynomial; that is
    $x_{p}(t)=b_{i}t^{i}$
for certain coefficients $b_{i}$.

### Exercises

* Exercise 3.3.1: Find all solutions to the ODE $x'(t)+2x(t)=2t^{2}+4t+3$. (Hint: What does theorem 3.1 say about the number of solutions to that problem with a given fixed initial condition?)

## Examples

Example 3.2: