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