Constant Parameters in Solution of ODEs

Question

Consider the ode $y'=y^2$. We know that $y=-1/(c+x)$ is a family of one parameter solutions. This is not general solution since the $y=0$ can not be obtained from it. Now consider the linear equation $y'=y$. We know that $y=ce^x$ is its general solution. For $c=0$ we have the trivial solution. Let us to obtain this general solution step by step:

$$ y'=y\rightarrow dy=ydx. $$ For $y\neq 0$ we have

$$ dy/y=dx\rightarrow \ln|y|=x+k\rightarrow y=\pm e^ke^x=ce^x. $$ where $c=\pm e^k$. There isn't any finite $c$ to obtain the solution $y=0$. We see that the coefficient of $e^x$ is nonzero. How we can find the solution $y=0$? The only way is to let $k\rightarrow -\infty$.

Question: In theory of differential equations we know that linear odes have a basis of linear independent solutions on an interval I which generates the space of solutions. The above example shows that for such a simple equation, we can not find all solutions. How we can introduce this ambiguity? Is it possible to let $k\rightarrow -\infty$ to find $c=0$? Note that before integration we supposed that $y\neq 0.$

Answer 1

Here is the thing: in modern expositions of the theory, it is insisted that $y$ be a function of $x$ and differential equations are written in "normal form" where the derivative (or the highest order derivative) is expressed as a function of the lower order derivatives and $x$. A more symmetric approach allows to see the whole picture. Namely, you write the equation in differential form $$ ydx-dy=0 $$ and multiply by the integrating factor $e^{-x}$, $$ e^{-x}ydx-e^{-x}dy=0 $$ or $$ d(e^{-x}y)=0 $$ which is exact and has as solutions all the level sets of $F(x,y)=e^{-x}y$, that is, all curves of the form $y=Ce^x$ for real $C$, in particular $y=0$.

BTW that is how mathematicians centuries ago used to proceed, but nowadays everything has to be a function (just a fashion).

In modern textbooks they tell you to find "special solutions" first (in this case the solution $y=0$ would be special) and then the "general" one, which is not that general because you divided the equation by $y$.