Use the method of variation of parameters to solve the differential equation

Question

d²y/dx²+9y=cosec²(3x).

Answer 1

To solve the given differential equation, we must first solve the associated differential equation d²y/dx²+9y = 0. Since d²y/dx²+9y = 0 is a homogeneous linear equation with constant coefficients, we know that the associated auxiliary equation is m² + 9 = 0. Since the auxiliary equation has m = ±3i as solutions, we know that y_c = c₁cos(3x) + c₂sin(3x) is the general solution to d²y/dx²+9y = 0.

We must now use the method of variation of parameters to solve the given differential equation.The preceding work tells us that y₁ = cos(3x), y₂ = sin(3x), and f(x) = csc²(3x). Since dy₁/dx = -3sin(3x) and dy₂/dx = 3cos(3x), we also know that

W = | cos(3x) sin(3x) | = 3cos²(3x) + 3sin²(3x) = 3,

| -3sin(3x) 3cos(3x) |

W₁ = | 0 sin(3x) | = 0 - csc²(3x)*sin(3x) = -(1/sin²(3x))*sin(3x) = -1/sin(3x) = -csc(3x),

| csc²(3x) 3cos(3x) |

W₂ = | cos(3x) 0 | = cos(3x)*csc²(3x) + 3sin(3x)*0 = cos(3x)*(1/sin²(3x)) = cos(3x)/sin²(3x),

| -3sin(3x) csc²(3x) |

du₁/dx = W₁/W = -csc(3x)/3 = -(1/3)csc(3x),

and du₂/dx = W₂/W = (cos(3x)/sin²(3x))/3 = (1/3)(cos(3x)/sin²(3x)).

Integrating du₁/dx with respect to x by using the substitution U = 3x and the general result ∫ csc(x) dx = ln|csc(x) - cot(x)| + C then gives us u₁ = -(1/9)ln|csc(3x) - cot(3x)|.

Integrating du₂/dx with respect to x by using the substitution U = sin(3x) then gives us u₂ = -(1/9)csc(3x). Thus, a particular solution to the original differential equation is y_p = -(1/9)ln|csc(3x) - cot(3x)|cos(3x) - (1/9)csc(3x)sin(3x) = -(1/9)ln|csc(3x) - cot(3x)|cos(3x) - 1/9. The general solution of the original differential equation is thus y = y_c + y_p = c₁cos(3x) + c₂sin(3x) - (1/9)ln|csc(3x) - cot(3x)|cos(3x) - 1/9.