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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?

Ambesh
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    Can you clarify with an example? – Git Gud Apr 06 '13 at 11:58
  • For example, you are solving a differential equation and in the process of finding the solution you assume that the the solution is differentiable, and you don't know that yet. So technically the process of deduction is wrong, even if the solution may be right. – Ambesh Apr 06 '13 at 12:05
  • When solving a differential equation such as $F(y,y',y'',x)=0$ the task is to find a function $y(x)$ such that $y'$ and $y''$ exist and $y,y', y'',x$ satisfy $F(y,y', y'',x)=0$. Since we are explicitly asked to find a twice differentiable function in this example, I see no unjustified assumption that $y$ is twice differentiable. – Hagen von Eitzen Apr 06 '13 at 12:44
  • Other solutions are often called the non-standard solutions. Those might be very, very hairy and it is best to keep away from them (and definitely not to poke with a stick) if you don't want to get burned. Nevertheless, there are some shields and magic items that let you fight these monsters (e.g. in differential equations world there are weak derivatives or Sobolev spaces), but I would recommend getting some exp-points first. However, with appropriate preparation, go ahead and slay your dragon! :-) – dtldarek Apr 06 '13 at 12:52
  • I'd still be happy with an explicit example of $F$ and $\mathscr A$ to see what the question really means. – Hagen von Eitzen Apr 06 '13 at 13:49

2 Answers2

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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.

Git Gud
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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.

Zander
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