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Assuming I have the following homogeneous ODE equation: $$a\cdot y'' + b\cdot y' + c \cdot y = 0$$ Why for $(b^2 - 4\cdot a\cdot c=0) \quad $,(meaning, when $m_1=m_2$) then the solution is: $$y = C_1\cdot e^{m_{1}x} + C_2\cdot x \cdot e^{m_{2}x}$$ Why isn't it simply: $$y = C_1\cdot e^{m_{1}x} + C_2 \cdot e^{m_{2}x}$$ ?
Also, why did they choose to multiply $C_2$ with $x$? Why not having a totally different approach for the solution when $(m1=m2)$ (e.g. diving the equation with $x$)?

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second order differential equation needs two independent solutions so that it can satisfy the two initial conditions like $y(a) = \alpha, y^\prime(a) = \beta.$ when the indicial equation has a repeating root, which happens when $b^2 = 4ac,$ you only get one solution $e^{rx}$

you find the other solution by pretending the repeating root is really two roots $r -\epsilon$ and $r + \epsilon$ coming together to be $r, r$ so that $${e^{(r+\epsilon)x} - e^{(r-\epsilon)x} \over 2 \epsilon }= {e^{rx} (e^{\epsilon x} - e^{-\epsilon x}) \over 2 \epsilon} ={e^{rx}[1 + \epsilon x + \cdots -(1 - \epsilon x + \cdots)] \over 2 \epsilon} = xe^{rx} \mbox{ as } \epsilon \to 0$$ is also a solution. we have also used the fact that the linear combinations of the two solutions $e^{(r+\epsilon)x}, e^{(r-\epsilon)x}$ is again a solution.

so we have now two solutions $e^{rx}$ and $xe^{rx}$ when the indicial $am^2 + bm + c = 0$ has a repeating root.

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  • $\begingroup$ Thanks! Could you please elaborate regarding the equation? I'm not sure how the $x$ got there.. $\endgroup$ – Dor Dec 27 '14 at 11:36
  • $\begingroup$ @Dor, i will elaborate on the limit of the quotient. $\endgroup$ – abel Dec 27 '14 at 14:05
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Maybe oversimplifying here but, for 2nd order ODEs you should have two unique solutions. If $m_1=m_2$, your two solutions would not be unique, multiplying by $x$ takes care of that. The other part (why you wouldn't divide) is that you actually want your solution to be a solution to your ODE!

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