y”-9y’+14y=3x^2-5cos2x+7xe^6x

To solve the given differential equation y'' - 9y' + 14y = 3x^2 - 5\cos(2x) + 7xe^{6x}, we will first find the complementary function by assuming y = e^{mx}.

Let's start by finding the complementary function:
Assume y = e^{mx}
Then, we have:
y'' = m^2 e^{mx},
y' = m e^{mx},
Substitute these into the differential equation:
m^2 e^{mx} - 9m e^{mx} +14e^{mx} = 0
m^2 - 9m + 14 = 0
(m-2)(m-7) = 0
m = 2,\, 7

Therefore, the complementary function is given by:
y_c = c_1 e^{2x} + c_2 e^{7x}

Next, we will find the particular solution for the given non-homogeneous part:
Given non-homogeneous part:
f(x) = 3x^2 - 5\cos(2x) + 7xe^{6x}
To find the particular solution, we can use the method of undetermined coefficients. We will assume the particular solution is in the form:
y_p = Ax^2 + Bx + C\cos(2x) + Dx^2e^{6x}
y_p' = 2Ax + B + 6Dx^2e^{6x} + Ccos(2x) - 2Csin(2x)
y_p'' = 2A + 12Dx^2e^{6x} - 4Ccos(2x) - 4Csin(2x)

Now substitute y_p, y_p', y_p'' back into the differential equation:
y_p'' - 9y_p' + 14y_p = 3x^2 - 5\cos(2x) + 7xe^{6x}
(2A + 12Dx^2e^{6x} - 4Ccos(2x) - 4Csin(2x)) - 9(2Ax + B + 6Dx^2e^{6x} + Ccos(2x) - 2Csin(2x)) + 14(Ax^2 + Bx + Ccos(2x) + Dx^2e^{6x}) = 3x^2 - 5\cos(2x) + 7xe^{6x}

Matching coefficients of terms with similar types of functions:
2A - 9B + 14A = 3
-4C - 9A - 14C = 0
-4C - 9B + 14B = 0
12D - 9D + 14D = 7

Solving these equations will give the values of A, B, C, and D.

After finding A, B, C, and D, the general solution will be:
y = y_c + y_p

y = c_1 e^{2x} + c_2 e^{7x} + Ax^2 + Bx + C\cos(2x) + Dx^2e^{6x}

\boxed{y = c_1 e^{2x} + c_2 e^{7x} + Ax^2 + Bx + C\cos(2x) + Dx^2e^{6x}}