Why isn't there a constant within the sinusoidals in solutions to 2nd Order ODE's

Question

In reading my partial differential equations textbook an ODE was solved in the following way: $$ \frac{d^2h}{dx^2}=\left(\frac{n\pi}{H}\right)^2 h \hspace{1cm} h = \sinh\left(\frac{n\pi(x-L)}{H}\right) $$ In this case, sinh can be reduced to the sum of sinh and cosh terms in $x$, not $(x-L)$, but it got me thinking about why we don't include a constant inside of sine and cosine functions, for example, in the general solution for negative eigenvalue: $$ \frac{d^2y}{dx^2} = -k^2 y \hspace{1cm} y=c_1\cos(kx+c_2) + c_3 \sin(kx+c_4) $$ This is my question - what, exactly, makes it unnecessary to include these additional constants $c_2$ and $c_4$?

Answer 1

In general, a second-order ODE should have a two-parameter family of solutions. In your second ODE you don't need $c_2$ and $c_4$ because with $c_1$ and $c_3$ you already have a two-parameter family. Indeed, expanding your sin and cos terms and rearranging gives you something of the same form without the $c_2$ and $c_4$:

$$c_1 \cos(kx + c_2) + c_3 \sin(kx + c_4) = b_1 \cos(kx) + b_2 \sin(kx) $$ where $$ b_1 = c_1 \cos(c_2) + c_3 \sin(c_4),\ b_2 = - c_1 \sin(c_2) + c_3 \cos(c_4) $$

As for your first equation, you have one parameter $L$, but you need another: you could write $$ y = c \sinh \left(\frac{n\pi (x-L)}{H}\right) $$