Gap in fundamental knowledge of differential equations

[oldstudent]

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.

 

[George Jones]

Both your solution and the book's solution to the homogeneous equation are correct, but the arbitary constants in the solutions are different, so it would be better to call your arbitrary constants A' and B'.

Do you see how the solutions are the same?

 

[HallsofIvy]

Since $cosh x= \frac{e^x+ e^{-x}}{2}$ and $sinh x= \frac{e^x- e^{-x}}{2}$, adding the two equations, $e^x= cosh x+ sinh x$, and subtracting $e^{-x}= cosh x- sinh x$. If you substitute those into $Ae^x+ Be^{-x}$ you get the book's solution with, as George Jones said, different constants.

By the way, cosh x and sinh x have the nice property that cosh 0= 1 and sinh 0= 0. Of course (cosh x)'= sinh x and (sinh x)'= cosh x. In particular, if y(x) is a solution to that equation, then y(x)= y(0)cosh x+ y'(0)sinh x.

 

[oldstudent]

Many thanks to you, George and Halls of Ivy, for your help.