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:

$$y(x)=c_1e^{2x}+e^{-x}(c_2\cos(\sqrt{3}x)+c_3\sin(\sqrt{3}x))$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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