A particular solution of the differential equation y"- 2y' + y = cosx is:
O yp = sinx / 2
O yp = cosx
O yp = -sinx / 2
O yp = cosx / 2
O yp = sinx

 y_p = - sin(x) / 2

Step-by-step explanation:

Given:

- The following ODE as such:

                                  y" - 2y' + y = cos(x)

Find:

The particular solution.

Solution:

- The particular solution resembles the non-homogeneous part of the ODE.

- So lets suppose we have a solution:

                                 y_p = A*cos(x) + B*sin(x)

Where, A is constant that needs to be determined.

- Differentiate the particular solution 2 times:

                                y'_p = - A*sin(x) + B*cos(x)

                                y"_p = -A*cos(x) - B*sin(x)

- Use the derivatives and plug the back in the ODE as follows:

   -A*cos(x) - B*sin(x) + 2*(A*sin(x) - B*cos(x)) + A*cos(x) + B*sin(x) = cos(x)

- Simplify and compare coefficients:

                               2A*sin(x)  - 2B*cos(x) = cos(x)  

We have:                 2A = 0     ,   -2B = 1

Hence,                     A = 0      ,   B = -1/2

- Hence we can write our particular solution to be:

                              y_p = - sin(x) / 2       Hence option C