How do limits work on solutions of differential equations? Can we compare the solutions of differential equations by taking limits with respect to a parameter?
For example, consider the following second-order ordinary linear differential equation:
$$y''(x) + b  y(x)' = 1$$
The solution will be an affine surface parametrized by $c_1$ and $c_2$, where given some initial conditions we can identify the unique solution:
$$y_{b}(x) = c_1 \frac{e^{-bx}}{b}+\frac{x}{b}+c_2$$
Now, if $b=0$, then the solution will be (parametrizing the affine space by $k_1$ and $k_2$):
$$y_{b=0}(x) = \frac{x^2}{2}+k_1x+k_2$$
I have two questions:


*

*For given $k_1$ and $k_2$ (or the corresponding initial condition), can we find $c_1$ and $c_2$, such that $y_{b}(x) \rightarrow y_{b=0}(x)$ as $b\rightarrow 0$?

*If not, what am I doing wrong? In particular, as I plot these functions for different values of $b$, I see a continuous transition between functions, but $b=0$ behaves like an anomaly.
Thanks!
 A: The continuity with respect to the parameter is under a fixed set of boundary conditions. Really these should be initial conditions. Boundary conditions can have discontinuities in certain cases, such as when a parameter in a linear problem with homogeneous boundary conditions passes through an eigenvalue.
For example, consider $y'+y=e^{kt},y(0)=0$, in a vicinity of $k=-1$. A particular solution for $k \neq -1$ is $\frac{e^{kt}}{1+k}$. The homogeneous solution is $Ce^{-t}$. The particular solution satisfies $y_p(0)=\frac{1}{1+k}$ so the solution to the IVP for $k \neq -1$ is
$$y=\frac{-e^{-t} + e^{kt}}{1+k}.$$
Sending $k \to -1$ in this expression (without having to know a particular solution for $k=-1$) gives the solution
$$y=te^{-t}$$
which is valid for $k=-1$.
A nice physical example of this is the problem
$$y''+y=\sin(kt),y(0)=0,y'(0)=0$$
with $k$ near $1$. This reveals that the unbounded amplitude oscillations of the problem $y''+y=\sin(t)$ arise not "discontinuously" but rather through arbitrarily large amplitude oscillations of $y''+y=\sin(kt)$ for $k$ near $1$.
