Question

Use the variation of parameters method to solve y′′+9y=0.25csc(3x)

Answer 1

To solve the differential equation y'' + 9y = 0.25csc(3x) using the variation of parameters method, you can follow these steps.

Step-by-step explanation:

To solve the differential equation y'' + 9y = 0.25csc(3x) using the variation of parameters method, we can follow these steps:

1. Find the complementary solution by solving the homogeneous equation y'' + 9y = 0. This gives us the solution yc(x) = A*cos(3x) + B*sin(3x), where A and B are constants.
2. Find the particular solution by assuming a variation of parameters of the form yp(x) = u(x)*cos(3x) + v(x)*sin(3x). Differentiate this equation twice to find y''p.
3. Substitute yp(x), y''p(x), and the given right-hand side into the original differential equation. This will give you two equations involving u''(x), v''(x), u(x), v(x), and the trigonometric terms.
4. Solve the resulting system of equations to find the expressions for u(x) and v(x) in terms of c.

The solution to the given differential equation is y(x) = yc(x) + yp(x), where yc(x) is the complementary solution and yp(x) is the particular solution.