Question

Show that the function u(x, t) = A(x + ct)^3 is a solution of the one-dimensional wave equation.

Answer 1

y = f( x , y) is a wave equation when
d²y/dx² =( constant) d²y/dt²
now,
u( x , t) =A( x + Ct)³
du/dx = 3A( x + Ct)² .1 =3A(x +Ct)²
again, differentiate wrt x
d²u/dx² = 6A( x +Ct) -------(1)

now , u(x , t) differentiate wrt t
du/dt =3A(x +C t)².C
d²u/dt² =6AC² ( x +Ct) -------(2)

from equation (1) and (2)

d²u/dx² ={6A/6AC²}d²y/dt²

d²y/dx² =(1/C²) d²y/dt²
here C is constant so,
this is in the form of
d²y/dx² =( constant) d²y/dt²

so, u(x , t ) = A( x + Ct)³ one -dimensional equation of wave .

Answer 2

One dimensional wave equation is :
     d²y/dx² = 1/v² d²y/dt²      --- (1)

where  v = velocity of the wave,
          x is the distance in the direction of propagation of the wave
          y = quantity : displacement, pressure, torsion, longitudinal vibration
          t = time

This is found by d'Alembert. This progressive transverse wave equation describes the vibrations of a string in a musical instrument, sound propagation and heat transfers. This is applicable for standing waves.

                     Given:    u(x, t) = A(x + c t)³   ---(2)
Partial derivatives:
     du(x,t)/dx = 3 A (x + ct)²           d²u(x,t)/dx² = 6 A (x+ ct)
     du(x,t)/dt = 3 c A (x + ct)²         d²u(x,t)/dx² = 6 c² A (x + ct)

Substitute them in (1):    d²u/dx² = 1/c² * d²u/dt²  = 6 A (x+ct)

Hence the given solution satisfies the 1-d wave equation.