Finding general solution of linear ODE inhomogeneous

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ohspyro89
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Homework Statement


Find the general solution to y''-2y'-24y=50e6x-14cos(x)-175sin(x)


Homework Equations


I can't figure out how to solve for B,C,D, and E. I'm wondering if I did something wrong.



The Attempt at a Solution



I'm attaching photos, since it'd take forever to type this all out. I'm stuck on this one, and I'm working on another problem at the same time.

Handwritten work on lined paper solving y'' - 2y' - 24y = 50e^6x - 14cos x - 175sin x: characteristic equation m^2 - 2m - 24 = 0 factored as (m-6)(m-4), giving complementary solution yc = C1e^6x + C2e^4x, then undetermined coefficients trial roots listed.

Handwritten notes on lined paper solving a linear ODE: characteristic roots m = 4, 6, 6i, 5i, a particular solution yp with coefficients A through E, and a system of equations matching coefficients of e^6x, xe^6x, sin x, cos x, x sin x and x cos x.

Handwritten work on lined paper: substituting a particular solution yp = Axe^6x + Bsin x + Ccos x + Dxsin x + Excos x and its derivatives into an ODE, with coefficient equations listed for e^6x, xe^6x, sin x, cos x, xsin x and xcos x, and partial values A = 10 and D = -12E noted.
 
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it should be factored as (m-6)(m+4)
not (m-6)(m-4) ill look at the rest.
 
Yes, you did something wrong. Your particular solution should be
yp = Axe6x + Bsin(x) + Ccos(x)

The general solution will be y = c1e6x + c2e-4x + yp
 
*Face-Palm*

Alright, I'll work on it. Thanks guys!

Also, was I right to get rid of one of the +-i for the undetermined coefficients part? So, now I don't have a D and E to worry about, but it covers the cosx-sinx.
 
You really don't need either of the +/- i pairs. If your roots of the char. equation include +/- i, the solutions will include e^(ix) and e^(-ix), but you can take a linear combination of these and work with cos(x) and sin(x) instead.
 
How's this? I got a constant to be zero, which seems odd to me. Is this the right general solution?

Handwritten working on lined paper solving a linear ODE: particular solution yp = Axe^6x + Bsinx + Ccosx, with coefficient equations and results A=10, B=7, C=0, ending in y = C1e^6x + C2e^-4x + 10xe^6x + 7sinx.
 
Looks reasonable - you can check it yourself. Is y'' - 2y' - 24y = 50e6x-14cos(x)-175sin(x) an identity? IOW, if you replace the 3 terms on the left with your solution and its first two derivatives, do you end up with what you have on the right?