MHB Paul's question at Yahoo Answers regarding a 3rd order linear homogeneous ODE

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The discussion centers on solving the third-order linear homogeneous ordinary differential equation (ODE) given by y''' - 8y = 0. The characteristic equation derived from this ODE is r^3 - 8 = 0, which factors to (r - 2)(r^2 + 2r + 4) = 0. The roots of the equation are identified as r = 2 and r = -1 ± i√3. Consequently, the general solution to the ODE is expressed as y(x) = c1e^(2x) + e^(-x)(c2cos(√3x) + c3sin(√3x)). This solution incorporates both real and complex roots, reflecting the behavior of the system described by the ODE.
MarkFL
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Here is the question:

Differential equations factoring?

Find the general solution to the following

y'''-8y=0

I have posted a link there to this topic so the OP can see my work.
 
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Hello Paul,

We are given to solve:

$$y'''-8y=0$$

The associated characteristic equation is:

$$r^3-8=(r-2)(r^2+2r+4)=0$$

Hence, the roots are:

$$r=2,\,-1\pm i\sqrt{3}$$

and so the solution is:

$$y(x)=c_1e^{2x}+e^{-x}(c_2\cos(\sqrt{3}x)+c_3\sin(\sqrt{3}x))$$
 
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