Most general solution of an ODE
Matthew Harrington 
Consider $y''(x)=0$. It seems obvious that the most general solution to this equation is given by $y=Ax+B$, where $A, B \in \mathbb{R}$, which can be confirmed through integration. But how do we know that this is indeed the most general solution to this equation, and that no other had been omitted? I know there are ODE's where some solution may be omitted when standard methods are used, e.g the trivial division by zero, so how do we know if a solution to an ODE is the most general one?
$\endgroup$ 32 Answers
$\begingroup$The answer is right in your question. Integration is the straight-forward way to prove it.
Assuming $y''(x) = 0$ as a differential equation over the real variable $x$ :
$$y''(x) = 0 \Rightarrow \int y''(x)dx =\int0dx\Rightarrow y'(x) = A \Rightarrow\int y'(x)dx = \int Adx \Rightarrow$$
$$\Rightarrow$$
$$y(x) = Ax + B \space , \space \text{where} \space A,B \space \text{constants} \space \in \mathbb R$$
$\endgroup$ $\begingroup$If a function's derivative is zero, then the function is constant; this is an essential point regarding integration. So you know that $y'(x)=A$ for some constant $A$. Now consider $z(x)=Ax$. Then $$ (z-y)'=z'-y'=A-A=0. $$ So $y(x)-z(x)=B$ for some constant $B$. That is, $y(x)=Ax+B$.
In general, for some nice enough differential equations one can prove existence and uniqueness results. So, for instance, for a second order linear, homogeneous, constant coefficient equation, one can prove that there is a unique solution provided that the value of the function and its derivative are prescribed at a single point. Then, if you find a solution to the initial value problem, you will know its the only one because of the theorem.
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