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I have posted a link there to this topic so the OP can see my work.Differential equations factoring?

Find the general solution to the following

y'''-8y=0

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- Thread starter
- Admin
- #1

I have posted a link there to this topic so the OP can see my work.Differential equations factoring?

Find the general solution to the following

y'''-8y=0

- Thread starter
- Admin
- #2

We are given to solve:

\(\displaystyle y'''-8y=0\)

The associated characteristic equation is:

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

Hence, the roots are:

\(\displaystyle r=2,\,-1\pm i\sqrt{3}\)

and so the solution is:

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