Show that `(a)y=(x+vt)^(2),(b)y=(x+t)^(2),(c )y=(x-vt)^(2)`, and (d) y=2 sin xcos vt are each a solution of one dimensional wave equation but not (e) y=x^(2)-v^(2)t^(2)` and (f) `y=sin 2x cos vt.`

Explanation

Differentiating expression (i) twice `w.r.t.t` we have,
`(del^(2)y)/(delx^(2)=(2)`
And differentiating expression (i) twice `w.r.t.x` we have,
`(del^(2)y)/(delx^(2)=2v^(2)`
clearly, `(del^(2)y)/(delt^(2))=v^(2)(del^(2)/delx^(2))`
Thus expression (i) is a solution of the one dimensional wave equation.
Similarly, treatment can be done for (ii), (iii) and (iv). note: the expression (iv) satisfies differential equation of wave, but it does not represent progressive wave. it represents stationary wave.
(v). Differentiating expression (v) twicee `w.r.t.t` we have
`(del^(2)y)/delt^(2)=-2v^(2)`
clearly `(del^(2)y)/(delt^(2)!=v^(2)(del^(2)y)/(delx^(2)`,
so the expression (v) is not a solution of the one dimensional wave equation.
(vi) Differentiating expression (vi) twice `w.r.t.t, we have `(del^(2)y)/(delt6(2))=-v^(2) sin2xcos vt=-v^(2)y`
`(del^(2)y)/(delx^(2)=-4y`
clearly `(del^(2)y)/(delt^(2)!=v^(2)(del^(2)y)/(delx^(2)` and therfore the expression is not a solution of the one dimensional wave equation.