Some advice/guidance required...
Given the following problem:
$$ y'' - t^2 y = a\cos(tx)+b\cosh(t x), $$ where $$ y(0) = 1, y(1) = 2 $$
The method I am using is as follows:
1) Solve the related homogeneous equation (gives $ y_c = A\cosh(tx) + B\sinh(tx) $)
2) Use a particular solution to generate the general solution ($ y_g = y_c + y_p $). The particular solution I am using is $$ y_p = C\cos(tx) + D\cosh(tx) $$
The question I have is, am I using the correct $ y_p $ ?
I don't get the solution I require so I suspect I'm using an incorrect $ y_p $ but I'm not sure why?
Also, firstly, as $ \cosh(t x) $ is used within $ y_c $ do I need to 'adjust' it within $ y_p $ with another constant to allow the Math to work? This would give $$ y_p = C\cos(tx) + D w \cosh(tx) $$ Where I have added the constant $w$.
Or... should I be taking $ a\cos(tx) $ and $ b\cosh(t x) $ seperately and solving for each to give $ y_{p1} $ and $ y_{p2} $ and then adding to get $ y_g = y_c + y_{p1} + y_{p2} $ ?
Thank you in advance for any pointers.