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            

a) 0.25x times 3. (could be wrong on equation.)

b) 2.25 pound of dark chocolate. (can i eat all the chocolate? plz? anyway, again, could be wrong, but i hope im right. all the chocolate together should be 5.25 pounds)

have a good day!

idk lol

step-by-step explanation: