y''-2.5y'=9.81
I can't seem to be able to solve this. Any help would be awesome.
This is a second order equation with constant linear coefficients. We have a method to use to solve problems of these types, where we assume the solution to our equation follows some form (your book should list this method as "Undetermined Coefficients").
The first step in solving a DE of this form is to find the homogenous solution to our DE:
$\displaystyle y_h:$ $\displaystyle y''-2.5y'=0$
Solve the auxillary equation:
$\displaystyle m^{2}-2.5m=0 \Rightarrow m(m-2.5)=0$
Thus the solutions to our homogenous equation will be:
$\displaystyle y_h: c_1+c_2e^{2.5x}$
We now need to find a PARTICULAR solution to our differential (the general solution to the differential will be a combination of the homogenous and particular solution). We can reason that the particular solution will be a constant function, since our differential is a constant (9.81).
If we make the assumption that $\displaystyle y_p = A$, we run into a problem - a constant solution is already a solution to the homogenous part of our differential. Thus we need to multiply through by "x" to get Ax as the form of our particular solution.
At this step you should know how to complete this type of problem.