Answer:
y_p = - sin(x) / 2
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