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## Homework Statement

y''' + 2y'' - 3y' = 0

I solved the above using the following, where D is the differential operator:

D(D - 1)(D + 3)y = 0

Let u(x) = (D + 3)y

Hence,

(D

^{2}- D)[u(x)] = 0. This 2nd Order in u(x), linear, homogenous with constant coefficients. Hence the solution given by characteristic equation is (from lambda = {0,1}):

u(x) = C1 + C2*exp(x)

Plugging this result back into the factorized DE:

(D + 3)y = C1 + C2*exp(x)

y' + 3y = C1 + C2*exp(x)

which can be solved by integration factor:

y(x) = (1/3)*C1 + (1/4)C2*exp(x) + C3*exp(-3x)

Checking with Wolfraalpha gives:

y(x) = (-1/3)*C1*exp(-3 x)+C2*exp(x)+C3

which looks similar to mine, but is a little off. Namely the negative sign and the fact that the factor of (1/3) is associated with an exponential and my factor of (1/3) is associated with a constant only.

Any thoughts on this?

Also: Would it be appropriate to simply write (1/3)*C1 as just plain old C1? that is, can't I just "absorb" all of the numerical coefficients into the parametric coefficients?

thanks!