This is a question where the answer is given but I don't understand how to derive it. The answer to the homogeneous eq. is: A cosh(x) + B sinh(x) with no explanation of how it is arrived at and the given answer then goes on to describe how the non-homogeneous part is solved. Diff. eq. : d^2y/dx^2 - y = x^2 My solution: the homogeneous eq. is: p^2 - 1 = 0 and since its the diff. of two squares the roots are +1 and -1 so my solution is; y (char) = Ae^x + Be^-x Equation 1 I know that sinh x = (e^x - e^-x)/2 & cosh x = (e^x + e^-x)/2 and I know how the complex root solutions are manipulated to give a solution in terms of the trig functions. I assumed from the aforementioned that it ought to be obvious to me how to transform Eq. 1 because the two "introductory" books I've looked at just say it's related to the eq. d^2y/dx^2 + y = 0. I guess since it doesn't have complex roots it wouldn't have occurred to me to modify the solution further and I haven't yet "seen" how to go about it. I would be most grateful if someone could explain.