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Suppose I have 3 equations:\begin{eqnarray} y=Ae^{m_1x}+Be^{m_2x}\\ y^\prime=Am_1e^{m_1x}+Bm_2e^{m_2x}\\ y^{\prime\prime}=A{m_1}^2e^{m_1x}+B{m_2}^2e^{m_2x} \end{eqnarray} Here $m_1$ and $m_2$ are fixed constants. I need to eliminate $A$ and $B$ from the above three equations. One thing I can do is to solve for $A$ and $B$ from the last two equations and replace their values in the first equation, but that's a very tedious process. I am looking for a simpler method. Any help appreciated. Thanks!

P.S: One thing I noticed is that the answer looks like this: $y^{\prime\prime}-(m_1+m_2)y^\prime+(m_1m_2)y=0$. Notice that this is somewhat similar to the polynomial equation: $x^2-(m_1+m_2)x+(m_1m_2)=0$
Where $m_1$ and $m_2$ are solutions for $x$.

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  • $\begingroup$ Why are you trying to do this? $\endgroup$
    – Ninad Munshi
    Mar 12, 2020 at 4:33
  • $\begingroup$ @NinadMunshi Just for fun $\endgroup$
    – Terrarium
    Mar 12, 2020 at 4:36
  • $\begingroup$ Your note is related to the fact on where these expressions come from. They come from solving for the roots of the associated characteristic equation of a linear constant coefficient differential equation. $\endgroup$
    – Ninad Munshi
    Mar 12, 2020 at 4:46
  • $\begingroup$ @NinadMunshi Actually I was trying to find the differential equation for $$y=Ae^{-5x}+Be^{-3x}$$. But I thought of finding the most general possible case and skip all the exercises that fit into that case. $\endgroup$
    – Terrarium
    Mar 12, 2020 at 7:58

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I suppose $m_1 \ne m_2$: $$(r-m_1)(r-m_2)=0$$ $$\implies r^2-(m_1+m_2)r+m_1m_2=0$$ Then you have the DE: $$y''-(m_1+m_2)y'+m_1m_2y=0$$

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  • $\begingroup$ Yes, that's exactly what peeked into my mind, but how does $r$ relate to my differential equations? $\endgroup$
    – Terrarium
    Mar 12, 2020 at 4:35

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