Getting the general solution to $G''(x) - c G(x) = 0$, for $(c\neq 0)$, in the form of hyperbolic/trigonometric functions

Question

I have this second order PDE: $$G''(x) - c G(x) = 0$$ where c is constant.

To find a general solution of this we have to consider three cases:

$1)$ $c=0$, then we have $G''(x) = 0$, then $G(x) = Ax + D$.

$2)$ $c > 0$, then if we let $G(x) = e^{rx}$, we will get general solution $G(x) = Ae^{\sqrt{c}\,x} + Be^{-\sqrt{c}\,x}$

$3)$ $c<0$, then we will get the general solution $G(x) = Ae^{i\sqrt{c}\,x} + Be^{-i\sqrt{c}\,x}$.

My question is how do we get in a case $(2)$ the general solution of the form

$$G(x) = E \cosh(\sqrt{c}\,x) + D \sinh(\sqrt{c}\,x)$$

and in case $(3)$

$$G(x) = E \cos(\sqrt{-c}\,x) + D \sin(\sqrt{-c}\,x)$$

Answer 1

Her the given equation is $$G''(x) - c G(x) = 0$$

Case $1:$ Let $c=0$, then the given equation becomes $$G''(x)=0$$ whose general solution is$$G(x)=Ax+B$$ where $A$ and $B$ are arbitrary constants.

Case $2:$ Let $c\lt 0$, i.e., $c=-m^2$

Then the given equation becomes

$G''(x) +m^2 G(x) = 0$

$\implies (D^2 +m^2)G(x)=0$ where $D\equiv \frac{d}{dx} $

Roots of the auxiliary equation $k^2 + m^2 = 0$ is $k= \pm i m=\pm i\sqrt c$

General solution of this equation is $$G(x)=a e^{i\sqrt c x} + b e^{-i\sqrt c x}$$ i.e, $$G(x)=A \cos(\sqrt c x)+B\sin(\sqrt c x)$$

Case $3:$ Let $c\gt 0$, i.e., $c=m^2$

Then the given equation becomes

$G''(x) -m^2 G(x) = 0$

$\implies (D^2 -m^2)G(x)=0$

Roots of the auxiliary equation $k^2 - m^2 = 0$ is $k= \pm m=\pm \sqrt c$

General solution of this equation is $$G(x)=a e^{\sqrt c x} + b e^{-\sqrt c x}$$i.e., $$G(x)=A \cosh(\sqrt c x)+B\sinh(\sqrt c x)$$

Note:

$e^{ix}=\cos x + i \sin x$

$e^{-ix}=\cos x - i \sin x$

$\cosh x +\sinh x = e^x$

$\cosh x -\sinh x = e^{-x}$

$\cos x = \frac{1}{2} (e^{ix}+e^{-ix})$

$\sin x = \frac{1}{2i} (e^{ix}-e^{-ix})$

$\cosh x = \frac{1}{2} (e^{x}+e^{-x})$

$\sinh x = \frac{1}{2} (e^{x}-e^{-x})$

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$G(x) = a e^{i\sqrt c x} + b e^{-i\sqrt c x}$

$\implies G(x)= a(\cos \sqrt c x + i \sin \sqrt c x)+b(\cos \sqrt c x - i \sin \sqrt c x)$

$\implies G(x)=(a+b)\cos \sqrt c x + (ia-ib)\sin \sqrt c x$

$\implies G(x)=A\cos \sqrt c x +B\sin \sqrt c x $

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$G(x) = a e^{\sqrt c x} + b e^{-\sqrt c x}$

$\implies G(x)= a(\cosh \sqrt c x + \sinh \sqrt c x)+b(\cosh \sqrt c x - \sinh \sqrt c x)$

$\implies G(x)=(a+b)\cosh \sqrt c x + (a-b)\sinh \sqrt c x$

$\implies G(x)=A\cosh \sqrt c x +B\sinh \sqrt c x $