One dimensional wave equation is

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  1. \(\frac{{{\partial ^2}y}}{{\partial {t^2}}} = {\alpha ^2}\frac{{{\partial ^2}y}}{{\partial {x^2}}}\)
  2. \(\frac{{\partial y}}{{\partial t}} = {\alpha ^2}\frac{{{\partial ^2}y}}{{\partial {x^2}}}\)
  3. \(\frac{{{\partial ^2}y}}{{\partial {t^2}}} = -\frac{{{\partial ^2}y}}{{\partial {x^2}}}\)
  4. None of these

Answer (Detailed Solution Below)

Option 1 : \(\frac{{{\partial ^2}y}}{{\partial {t^2}}} = {\alpha ^2}\frac{{{\partial ^2}y}}{{\partial {x^2}}}\)

Detailed Solution

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Concept:

Wave equation:

It is a second-order linear partial differential equation for the description of waves (like mechanical waves).

The Partial Differential equation is given as, 

\(A\frac{{{\partial ^2}u}}{{\partial {x^2}}} + B\frac{{{\partial ^2}u}}{{\partial x\partial y}} + C\frac{{{\partial ^2}u}}{{\partial {y^2}}} + D\frac{{\partial u}}{{\partial x}} + E\frac{{\partial u}}{{\partial y}} = F\)

  B2 – 4AC < 0

   Elliptical

  2-D heat equation

  B2 – 4AC = 0

   Parabolic

  1-D heat equation

  B2 – 4AC > 0

   Hyperbolic

  1-D wave equation

For One-Dimensional equation,

\(α^2\frac{{{\partial ^2}y}}{{\partial {x^2}}} = \frac{{{\partial ^2}y}}{{\partial {t^2}}}\)

where, A = α2, B = 0, C = -1

Put all the values in equation (1)

∴ 0 - 4 (α2)(-1)

> 0.

So, this is a one-dimensional wave equation.

Additional Information

\(\frac{{\partial y}}{{\partial t}} = {α ^2}\frac{{{\partial ^2}y}}{{\partial {x^2}}}\)

having A = α2, B = 0, C = 0

Put all the values in equation (1), we get 

0 - 4(α2)(0) = 0, therefore it shows parabolic function.

So, this is a one-dimensional heat equation.

\(\frac{{{\partial ^2}y}}{{\partial {t^2}}} = - \frac{{{\partial ^2}y}}{{\partial {x^2}}}\)

having A = 1, B = 0, C = 1

Put all the values in equation (1), we get 

0 - 4(1)(1) = -4, therefore it shows elliptical function.

So, this is a two-dimensional heat equation.

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