Answer
$f_1(z, t) = A e^{-b(z-vt)^2} $, $f_2(z, t) = A\sin [b(z-vt)]$ and $f_3(z, t) = \frac{A}{b(z-vt)^2+1}$
satisfies the wave equation whereas
$f_4(z, t) = A e^{-b(bz^2+vt)} $ and $f_5(z, t) = A \sin(bz)\cos(bvt)^3 $
doesn't satisfy the wave equation.
Work Step by Step
Wave equation is given by:
$$ \frac{\partial^2 f}{\partial z^2} = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} $$
For $f_1(z, t) = A e^{-b(z-vt)^2} $
$$ \frac{\partial f_1}{\partial z} = -2Ab(z-vt)e^{-b(z-vt)^2} $$
$$ \frac{\partial^2 f_1}{\partial z^2} = -2Abe^{-b(z-vt)^2}[(-2b)(z-vt)^2+1]\qquad ...(1)$$
$$ \frac{\partial f_1}{\partial t} = 2Abv(z-vt)e^{-b(z-vt)^2} $$
$$ \frac{\partial^2 f_1}{\partial t^2} = -2Abv^2e^{-b(z-vt)^2}[(-2b)(z-vt)^2+1]\qquad ...(2)$$
From equation $(1) \text{ and } (2)$,
$$ \frac{\partial^2 f_1}{\partial z^2} = \frac{1}{v^2}\frac{\partial^2 f_1}{\partial t^2} $$
Hence $f_1$ satisfies the wave equation.
For $f_2(z, t) = A\sin [b(z-vt)]$
$$\frac{\partial f_2}{\partial z} = Ab \cos[b(z-vt)]$$
$$\frac{\partial^2 f_2}{\partial z^2} = -Ab^2 \sin[b(z-vt)] \qquad ...(3)$$
$$\frac{\partial f_2}{\partial t} = -Abv \cos[b(z-vt)]$$
$$\frac{\partial^2 f_2}{\partial t^2} = -Ab^2v^2 \sin[b(z-vt)]\qquad ...(4)$$
From equation $(3) \text{ and } (4)$,
$$ \frac{\partial^2 f_2}{\partial z^2} = \frac{1}{v^2}\frac{\partial^2 f_2}{\partial t^2} $$
Hence, $f_2$ satisfies wave equation.
For $f_3(z, t) = \frac{A}{b(z-vt)^2+1}$
$$\frac{\partial f_3}{\partial z} = \frac{-2Ab(z-vt)}{(b(z-vt)^2+1)^2}$$
$$\frac{\partial^2 f_3}{\partial z^2} = \frac{-2Ab[1-3b(z-vt)^2]}{(1+b(z-vt)^2)^3} \qquad ...(5)$$
$$\frac{\partial f_3}{\partial t} = \frac{2Abv(z-vt)}{(b(z-vt)^2+1)^2}$$
$$\frac{\partial^2 f_3}{\partial t^2} = \frac{-2Abv^2[1-3b(z-vt)^2]}{(1+b(z-vt)^2)^3}\qquad ...(6)$$
From equation $(5) \text{ and } (6)$,
$$ \frac{\partial^2 f_3}{\partial z^2} = \frac{1}{v^2}\frac{\partial^2 f_3}{\partial t^2} $$
Hence, $f_3$ satisfies wave equation.
For $f_4(z, t) = A e^{-b(bz^2+vt)} $
$$ \frac{\partial f_4}{\partial z} = -2Ab^2ze^{-b(bz^2+vt)} $$
$$ \frac{\partial^2 f_4}{\partial z^2} = -2Ab^2e^{-b(bz^2+vt)}(1-2b^2z^2)\qquad ...(9)$$
$$ \frac{\partial f_4}{\partial t} = -2Abve^{-b(bz^2+vt)} $$
$$ \frac{\partial^2 f_4}{\partial t^2} =2Ab^2v^2e^{-b(bz^2+vt)}\qquad ...(10)$$
Clearly from equation $(9) \text{ and } (10)$,
$$ \frac{\partial^2 f_5}{\partial z^2} \ne \frac{1}{v^2}\frac{\partial^2 f_4}{\partial t^2} $$
Hence $f_4$ does not satisfy the wave equation.
For $f_5(z, t) = A \sin(bz)\cos(bvt)^3 $
$$ \frac{\partial f_5}{\partial z} = Ab\cos(bz)\cos(bvt)^3 $$
$$ \frac{\partial^2 f_5}{\partial z^2} = -Ab^2\sin(bz)\cos(bvt)^3\qquad ...(11)$$
$$ \frac{\partial f_5}{\partial t} = -3Abv(bvt)^2\sin(bz)\sin(bvt)^3 $$
$$ \frac{\partial^2 f_5}{\partial t^2} = -3Ab^2v^2(bvt)\sin(bz)[3(bvt)^3\cos(bvt)^3+2\sin(bvt)^3]\qquad ...(12)$$
Clearly from equation $(11) \text{ and } (12)$,
$$ \frac{\partial^2 f_5}{\partial z^2} \ne \frac{1}{v^2}\frac{\partial^2 f_4}{\partial t^2} $$
Hence $f_5$ does not satisfy the wave equation.
