Logical errors in math deductions Sometimes in mathematics we do this a lot:
Suppose that to find a function $y_1(x)$ that satisfies some equation (any type of equation, differential or whatever..):
$$F(y_1(x))=0$$
In order to find the solution we need to apply to the last equation properties that require that $y_1(x)\in{\mathscr{A}}$, being $\mathscr{A}$ a particular class of functions. But we don't know $y_1(x)$ so we don't know if it's an $\mathscr{A}$ function.
We assume that $y_1(x)\in{\mathscr{A}}$ and find a solution that indeed is in $\mathscr{A}$.
But how we do justify the method we used to find the solution. Was it right?
 A: Suppose you're given the differential equation: $y'+ay=b$ where $a,b$ are continuous functions on a certain interval $I$.
Let $F$ be the set of real differentiable functions defined on $I$.
If they ask you to find the solutions of this differential equation, what they're actually asking you to do is to prove that the set of solutions $S=\left\{y\in F:\bigl(\forall x\in I\bigr)\left(y'(x)+a(x)y(x)=b(x)\right)\right\}$ of the diffential equation is what you expect it to be, that is: $\left\{y\in F: \bigl(\exists A\in F\bigr)\bigl(\exists C\in \Bbb R\bigr)\bigl(\forall x\in I\bigr)\left(A'=a \wedge y(x)=e^{-A(x)}\int e^{A(x)}b(x)\,dx+Ce^{-A(x)}\right)\right\}$, i.e.:
$$S=\left\{y\in F\colon \bigl(\exists A\in F\bigr)\bigl(\exists C\in \Bbb R\bigr)\bigl(\forall x\in I\bigr)\left(A'=a \wedge y(x)=e^{-A(x)}\int e^{A(x)}b(x)\mathrm dx+Ce^{-A(x)}\right)\right\}$$
'Formalizing' things this way will get around those issues.

Most of the times the process of finding solutions to an equation is not a rigorous one. What you can do is allow yourself some liberty to find a possible set of solutions, then get back and formalize things as in the example above.
A: You could think of it as considering cases,
Case 1: $y\in \mathscr{A}$
Case 2: $y\not\in\mathscr{A}$
First you consider Case 1, then if you find a valid $y$ and the question is of the form "Find any such function" then you are done and don't need to consider Case 2, and there was no breach of logic.
If you fail to find a solution in Case 1 then it would be a logical error to declare that there are no solutions without considering Case 2.
