The general solution of the differential equation 3y'' - 6y' + 6y = e^x is found by combining the complementary solution of its homogeneous counterpart and a particular solution suited for the nonhomogeneous term.
To find the general solution of the differential equation 3y'' - 6y' + 6y = ex, we first look for the complementary solution (yc) by solving the associated homogeneous equation 3y'' - 6y' + 6y = 0. We then use the method of undetermined coefficients to find a particular solution (yp) to the nonhomogeneous equation. The general solution is the sum of both, ygeneral = yc + yp.
For the homogeneous equation, we find the associated characteristic equation, which is r2 - 2r + 2 = 0. This leads to complex roots, r = 1 ± i. This implies that yc = ex(C1cos(x) + C2sin(x)). To find yp, we assume a solution of the form yp = Aex and substitute back into the original equation to find A.
So, the answer involves both the complementary solution and the particular solution, yielding a general solution that accounts for the presence of the nonhomogeneous term ex.