A question on my homework: Solve the DE using undetermined coefficients. y''' - 5y'' + 6y' = 8+2sinx Ok, supposedly easy. I use a homogenous equation to solve for yc. y''' -5y'' +6y' = 0 m3-5m2+6m=0 m=0, -1, 6 yc = c1 + c2e-x+c3e6x Then use a trial solution to find yp, but this is where I get confused... g(x) = 8+2sinx yp = A+Bcosx+Csinx Right...? yp' = -Bsinx+Ccosx yp'' = -Bcosx-Csinx yp''' = Bsinx-Ccosx Bsinx-Ccosx+5Bcosx+5Csinx-6Bsinx+6Ccosx = 8+2sinx (-5B+5C) = 2 (5C+5B) = 0 B= -1/5 C=1/5 Here where I run into a problem -- because there is a constant in g(x), but only yp's derivatives are being added together, there is no way to make is equal 8+2sinx, or at least not as far as I can see. Would I just make A=8?