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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))
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Answer:
Step-by-step explanation:
Step 1: Start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers. Step 2: Write the number as a product of prime numbers.