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The question is quite simple but I am having doubt :- I am asked to find out the general solution of the differential equation $(D^2- 4)y=0$ where $D=\frac{dy}{dx}$.

If I want to use hyperbolic function then I have its general solution to be

$y(x)= a \cosh 2x + b \sinh 2x$ where $a$ and $b$ are constants.

Alternatively can I write the solution to be $y = ae^{2x}+ b e^{-2x}$ ? i.e using distinct and real roots $-2$ , $2$ ?

What relation does these two solution possess?

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Remember $\cosh x=\frac{e^{x}+e^{-x}}{2}$ & $\sinh x=\frac{e^{x}-e^{-x}}{2}$

Yes, you can write the general solution of $(D^2-4)y=0$ as follows $$y=C.F.(\text{complementary function}) +P.I. (\text{particular integral})$$

$$C.F.=ae^{2x}+be^{-2x}$$ since the particular integral is $0$. hence, the solution is
$$y=ae^{2x}+be^{-2x}+0$$$$=ae^{2x}+be^{-2x}$$

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