Recall that you started with " ". With , that will give . That second constant of integration, gives you the solution you already had.

Yes, because then and you already have [tex]e^{-2x} as a solution so your book is just not bothering to write that again. Both methods are correct.the book says u = c1e^(4x).

A little lost where? The general theorem for linear non-homogeneous equation is that if Y(x) is the general solution to the associated homogeneous equation and y(x) is any single solution to the entire equation, then Y(x)+ y(x) is the general solution to the entire equation. Having found the general solution to the associated homogeneous equation and the a specific solution to the entire equation, add them.the second solution is y2 = e^(-2x)e^(4x)=e^2x and yp = -1/2

so that y = c1e^(-2x) + c2e^(2x) - 1/2.

i know with a homogeneous equation that the general solution is y = yc + yp

jus a lil lost on this one..any help would be appreciated.

thanks in advance