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$$D(D-3)(D+4)[y]=0\;,$$ where $D$ is the differential operator, how to get the general solution of $y$? The solution suggest that it is $$y = 6c_1-2c_2\exp(-4t)+3c_3\exp(3t)$$

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Given the arbitrary $6$, $-2$ and $3$ in front of the constants, the exercise might want you to solve three successive differential equations - first $Du=0$, then $(D-3)v=u$ and finally $(D+4)w=v$.

This process can be circumnavigated with a few observations. First, the differential equation is third-order and thus has three linearly independent solutions. Second, the three operators $D$, $D-3$ and $D+4$ all commute. Third, $D+a$ annihilates $0$ for any constant $a$ (that is, $D+a$ applied to the zero function returns the zero function). Taken together, this implies that any solution to one of

$$\begin{cases} (D+0)\phi =0 && (i) \\ (D-3)\phi=0 && (ii) \\ (D+4)\phi=0 && (iii) \end{cases} $$

is also a solution to the original differential equation. To see this for e.g. (i), take a solution to the differential equation $D\phi=0$ and observe $$D(D-3)(D+4)\phi=(D-3)(D+4)D\phi=(D-3)(D+4)\,0=(D-3)\,0=0.$$

Finally, find the general solutions to (i), (ii) and (iii) and observe they are all linearly independent, and thus form a basis for the solution space we are after.

Remark. If a polynomial $p(\cdot)$ has repeated roots, the solution to $p(D)y=0$ is not so easy.

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the point is i don't know how they get the coefficient 6,-2,3 –  Mathematics Apr 11 '12 at 3:27
    
@Mathematics: It doesn't look to me like they got 6,-2,3 from anywhere. It's possible the exercise wants readers to understand and reference the equivalence of different forms of general solutions. That is, it is useful to see that a general solution of the form $$\alpha+\beta e^{-4t}+\gamma e^{3t}$$ is equivalent to the solution given, by a change-of-constant $$\alpha=6c_1,~\beta=-2c_2,~\gamma=3c_3.$$ Then again I might be missing something. In any case, I'll write out the long way too. –  anon Apr 11 '12 at 3:36
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Here's the long way. Write $D(D-3)(D+4)y=D(D-3)v=Du$. Solve successive equations:

$$\begin{array}{c l} Du & =0 \\ u & =A \\ \hline (D-3)v & = u \\ e^{-3t}(D-3)v & = Ae^{-3t} \\ D(e^{-3t}v) & =Ae^{-3t} \\ e^{-3t}v & = -(A/3)e^{-3t}+B \\ v & = -A/3+Be^{3t} \\ \hline (D+4)y & = -A/3+Be^{3t} \\ e^{4t}(D+4)y & = -(A/3)e^{4t}+Be^{7t} \\ D(e^{4t}y) & = -(A/3)e^{4t}+Be^{7t} \\ e^{4t}y & =-(A/12)e^{4t}+(B/7)e^{7t}+C \end{array}$$

$$\begin{array}{c l} y(t) & = -\frac{A}{12}+\frac{B}{7}e^{3t}+Ce^{-4t} \\ & =\alpha +\beta e^{3t}+\gamma e^{-4t}. \end{array}$$

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Okay. So For this you will have to do the ridiculous. I'm assuming that this is for an introductory mathematical physics course or something like that. Anyways. Just do it one at a time. I would suggest writing $(D-3)(D+4)g(x)=\frac{1}{D} 0$, then conintinuing.

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