Find a particular solution of?

Question

Find a particular solution of y"-2y'+2y=e^(x)[(2-2x-6x^2)cosx+(2-10x+6x^2)sinx].

 

How to find the correct initial guess for this problem?

Answer 1

I would start with y = a(x) e^x sin x + b(x) e^x cos x, just based on the form of the right hand side.

 

y' = [a(x) + a'(x) - b(x)] e^x sin x + [a(x) + b(x) + b'(x)] e^x cos x

 

y'' = [2a'(x) + a''(x) - 2b(x) - 2b'(x)] e^x sin x + [2a(x) + 2a'(x) + 2b'(x) + b''(x)] e^x cos x

 

y'' - 2y' + 2y = [a''(x) - 2b'(x)] e^x sin x + [2a'(x) + b''(x)] e^x cos x

 

Thus we will be looking for functions a(x) and b(x) such that

 

(1): a''(x) - 2b'(x) = 2 - 10x + 6x²

(2): 2a'(x) + b''(x) = 2 - 2x - 6x²

 

Differentiating (2) and then isolating a''(x) gives

 

a''(x) = -b'''(x)/2 - 1 - 6x

 

Plugging this into (1) and simplifying gives

 

b'''(x) + 4b'(x) = -6 + 8x - 12x²

 

We can easily get a particular solution for b(x) (ignoring integration constants):

 

b''(x) + 4 b(x) = -6x + 4x² - 4x³

 

b(x) = -1/2 + x² - x³

 

plug this into (2) and solve for a(x)

 

b''(x) = 2 - 6x

 

2a'(x) + 2 - 6x = 2 - 2x - 6x²

 

a'(x) = 2x - 3x²

 

a(x) = x² - x³

 

Thus a particular solution is

 

y = e^x [(x² - x³) sin x + (-1/2 + x² - x³) cos x]

 

 

Note: The homogeneous equation y_h'' - 2y_h' + 2y_h = 0 has solution

 

y_h = e^x (A sin x + B cos x).

 

Thus the coefficient -1/2 above can be absorbed into the coefficient B of the homogeneous component.

 

Therefore the best solution for the particular part is actually

 

y_p = e^x (x² - x³) (sin x + cos x)

 

 

To make sure that this is correct, I verified this solution with Mathematica.