# How to eliminate arbitrary constants from differential equations?

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 $$x$$.

• Why are you trying to do this? Mar 12, 2020 at 4:33
• @NinadMunshi Just for fun Mar 12, 2020 at 4:36
• 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. Mar 12, 2020 at 4:46
• @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. Mar 12, 2020 at 7:58

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: $$y''-(m_1+m_2)y'+m_1m_2y=0$$
• Yes, that's exactly what peeked into my mind, but how does $r$ relate to my differential equations? Mar 12, 2020 at 4:35