I am dumbfounded as to how to solve this particular equation:

Given $y(x+h)=y(x)-h(\alpha y(x)+\beta), y(0)=n$, where $\alpha$, $\beta$ and $n$ are constants, find $\lim_{h\to0}y(x)$

I tried solving this graphically, but nothing worked. I arrived at two answers in two trials: $y(x)=\beta x+ne^{-\alpha x}$ and $y(x)=\frac{\beta}{\alpha}+(n-\frac{\beta}{\alpha})e^{-\alpha x}$. Neither are correct.

I know nothing about differential equations, so if someone could clue me in, please do.

  • $\begingroup$ If you kow the answer, is it constant? $\endgroup$ – Tahir Imanov Jul 5 '16 at 12:53
  • $\begingroup$ If $x$ doesn't depend on $h$, then $\lim_{h\to 0}y(x) = y(x)$. $\endgroup$ – user137731 Jul 5 '16 at 13:50
  • $\begingroup$ $x$ does not depend on $h$, but $y(x)$ does. $\endgroup$ – Yoshimaster Jul 5 '16 at 13:52
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    $\begingroup$ How? The notation $y(x)$ means $y$ depends (solely) on $x$. If $x$ doesn't depend on $h$, then $y(x)$ shouldn't either. Do you mean $\lim_{h\to 0}y(x+h)$? $\endgroup$ – user137731 Jul 5 '16 at 13:53
  • $\begingroup$ I guess it's more of a sequence with a step size that approaches zero. I wish to find the function that this sequence approaches. $\endgroup$ – Yoshimaster Jul 5 '16 at 15:32

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