3y''-6y'+6y=e^x*secx

Solve - 6 (dy (x)) / (dx) + 3 (d^2 y (x)) / (dx^2) + 6 y (x) = e^x sec (x):

The general solution will be the sum of the complementary solution and particular solution. Find the complementary solution by solving 3 (d^2 y (x)) / (dx^2) - 6 (dy (x)) / (dx) + 6 y (x) = 0:

Assume a solution will be proportional to e^ (λ x) for some constant λ. Substitute y (x) = e^ (λ x) into the differential equation:

3 (d^2) / (dx^2) (e^ (λ x)) - 6 (d) / (dx) (e^ (λ x)) + 6 e^ (λ x) = 0

Substitute (d^2) / (dx^2) (e^ (λ x)) = λ^2 e^ (λ x) and (d) / (dx) (e^ (λ x)) = λ e^ (λ x):

3 λ^2 e^ (λ x) - 6 λ e^ (λ x) + 6 e^ (λ x) = 0

Factor out e^ (λ x):

(3 λ^2 - 6 λ + 6) e^ (λ x) = 0

Since e^ (λ x) !=0 for any finite λ, the zeros must come from the polynomial:

3 λ^2 - 6 λ + 6 = 0

Factor:

3 (2 - 2 λ + λ^2) = 0

Solve for λ:

λ = 1 + i or λ = 1 - i

The roots λ = 1 ± i give y_1 (x) = c_1 e^ ((1 + i) x), y_2 (x) = c_2 e^ ((1 - i) x) as solutions, where c_1 and c_2 are arbitrary constants. The general solution is the sum of the above solutions:

y (x) = y_1 (x) + y_2 (x) = c_1 e^ ((1 + i) x) + c_2 e^ ((1 - i) x)

Apply Euler's identity e^ (α + i β) = e^α cos (β) + i e^α sin (β) : y (x) = c_1 (e^x cos (x) + i e^x sin (x)) + c_2 (e^x cos (x) - i e^x sin (x))

Regroup terms:

y (x) = (c_1 + c_2) e^x cos (x) + i (c_1 - c_2) e^x sin (x)

Redefine c_1 + c_2 as c_1 and i (c_1 - c_2) as c_2, since these are arbitrary constants:

y (x) = c_1 e^x cos (x) + c_2 e^x sin (x)

Determine the particular solution to 3 (d^2 y (x)) / (dx^2) + 6 y (x) - 6 (dy (x)) / (dx) = e^x sec (x) by variation of parameters:

List the basis solutions in y_c (x):

y_ (b_1) (x) = e^x cos (x) and y_ (b_2) (x) = e^x sin (x)

Compute the Wronskian of y_ (b_1) (x) and y_ (b_2) (x):

W (x) = left bracketing bar e^x cos (x) | e^x sin (x)

(d) / (dx) (e^x cos (x)) | (d) / (dx) (e^x sin (x)) right bracketing bar = left bracketing bar e^x cos (x) | e^x sin (x)

e^x cos (x) - e^x sin (x) | e^x cos (x) + e^x sin (x) right bracketing bar = e^ (2 x)

Divide the differential equation by the leading term's coefficient 3:

(d^2 y (x)) / (dx^2) - 2 (dy (x)) / (dx) + 2 y (x) = 1/3 e^x sec (x)

Let f (x) = 1/3 e^x sec (x):

Let v_1 (x) = - integral (f (x) y_ (b_2) (x)) / (W (x)) dx and v_2 (x) = integral (f (x) y_ (b_1) (x)) / (W (x)) dx:

The particular solution will be given by:

y_p (x) = v_1 (x) y_ (b_1) (x) + v_2 (x) y_ (b_2) (x)

Compute v_1 (x):

v_1 (x) = - integral (tan (x)) / 3 dx = 1/3 log (cos (x))

Compute v_2 (x):

v_2 (x) = integral1/3 dx = x/3

The particular solution is thus:

y_p (x) = v_1 (x) y_ (b_1) (x) + v_2 (x) y_ (b_2) (x) = 1/3 e^x cos (x) log (cos (x)) + 1/3 e^x x sin (x)

Simplify:

y_p (x) = 1/3 e^x (cos (x) log (cos (x)) + x sin (x))

The general solution is given by:

Answer: y (x) = y_c (x) + y_p (x) = c_1 e^x cos (x) + c_2 e^x sin (x) + 1/3 e^x (cos (x) log (cos (x)) + x sin (x))