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Short example: consider the differential equation \begin{align*} f'(x)=\frac{k^2}{k^2+k+1}xf(x) \end{align*} where $k$ is a parameter. Wolfram Alpha tells me that the solution to this equation is \begin{align*} f(x)=ce^\frac{k^2 x^2}{2(k^2+k+1)} \end{align*} If I then take the limit as $k\rightarrow \infty$, the solution converges to \begin{align*} f(x)=ce^\frac{x^2}{2}. \end{align*} This is the same solution as if I had simply taken the limit before I tried to solve the differential equation, that is I instead solved \begin{align*} f'(x)=xf(x). \end{align*}

My question is, then, when is it appropriate to take a such a limit of a parameter BEFORE solving the differential equation. I haven't been able to find many references to this question. One book I found online (Ritger and Rose - "Differential Equations with Applications," p. 69) claims that "taking the limit of a parameter in a differential equation and then solving the differential equation is not always the same as solving the differential equation and then taking the limit of the parameter" but doesn't give references or conditions.

Any help (even just pointing me to the appropriate reference) would be appreciated!

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    $\begingroup$ The property you are interested in is called continuity of solutions with respect to parameter. Maybe that will help you in looking for it. I've had it covered in my analysis 3 course but I don't have my notes now :-( $\endgroup$ – Blazej Jul 27 '15 at 22:19
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Have a look at $$\frac{dy}{dx}=\frac{y^k}{x}$$ for an example where you really can't swap orders of taking the limit. For one order the answer is $$y = \ln x$$ and for the other order it is a peculiar expression which depends on whether $x$ is less than or greater than $1$.

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