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In Second-Order Linear ODE , we said ; $$y'' + A*y' + B*y=0$$. Let substitute $y=e^{rt}$ and then solution will be $$y=c_1 e^{r_1 t} + c_2 e^{r_2 t} \\$$ where $r_1\quad and \quad r_2$ is solution to $r^2 + Ar + B \\$ and $r_1 \ne r_2$. But we can easily see that $r_1$ and $r_2$ are also solution to the DE individually($y=e^{r_1t}\quad and \quad y=e^{r_2t}$).Therefore , why we use the form which is $y=c_1 e^{r_1 t} + c_2 e^{r_2 t}$

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  • $\begingroup$ You mean you'd use the solution $y = r_1t + r_2t$ as a solution? $\endgroup$ – Aldon Feb 1 '16 at 8:51
  • $\begingroup$ No , I didn't mention such a solution. $\endgroup$ – onurcanbektas Feb 1 '16 at 8:52
  • $\begingroup$ @Aldon I'm sorry ,I wasn't clear but I added a example in question.I hope it is enough. $\endgroup$ – onurcanbektas Feb 1 '16 at 8:56
  • $\begingroup$ @Leth: You need to add a comment that you assume $r_1 \not = r_2$ or treat the case of identical roots separately. $\endgroup$ – Carl Christian Feb 1 '16 at 8:57
  • $\begingroup$ @CarlChristian I added to the question. $\endgroup$ – onurcanbektas Feb 1 '16 at 9:00
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The point of the expression $y=c_1 e^{r_1 t} + c_2 e^{r_2 t}$ is that every solution can be represented by it.

It, for example, tells you that the solutions form a two-dimensional vector space, because if you sum two such expression, you get an expression of the same form (but with different $c_1$, $c_2$).

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