Explain $y''-y=0$

Question

I've solved this equation and got $c_0\cosh x + c_1\sinh x$. However, I've noticed that if the arbitrary constant $c_0$ doesn't equal $c_1$, this wouldn't work; you would have to convert the $\cosh x $and $\sinh x$ into solutions of $e^x$ first. Why is this the case, because in other differential equations, the arbitrary constants can be any number?

Answer 1

Why do you claim that it doesn't work? $$(c_0\cosh +c_1\sinh )''=c_0(\sinh)'+c_1(\cosh)'=c_0\cosh+c_1\sinh$$

Answer 2

It works for any $(c_0,c_1)$. Since $$(\cosh(x))''=\cosh(x)$$ and $$(\sinh(x))''=\sinh(x),$$ if $$y=c_0 \cosh(x)+c_1\sinh(x),$$ then $$y''=c_0\cosh(x)+c_1\sinh(x)=y,$$ and so $$y''-y=0.$$

It is also true that $\cosh(x)$ and $\sinh(x)$ are L.I. functions, since the wronskian determinant is not zero, because $$W(\cosh(x),\sinh(x))=\left|\begin{matrix}\cosh(x)&\sinh(x)\\(\cosh(x))'&(\sinh(x))'\\\end{matrix}\right|=$$ $$=\left|\begin{matrix}\cosh(x)&\sinh(x)\\\sinh(x)&\cosh(x)\\\end{matrix}\right|=\cosh^2(x)-\sinh^2(x)=1,\quad \forall x.$$

This implies that all the solutions are of the form $$y=c_0 \cosh(x)+c_1\sinh(x).$$

Answer 3

$\cosh x$ and $\sinh x$ are two independent linear combinations of $e^x$ and $e^{-x}$. Hence any linear combination of the two former can be expressed as a linear combination of the two latter, and conversely.

The hyperbolic functions are exponentials in disguise.