Question

3y´´-6y´+6y=e^x*secx

Answer 1

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))