Out of the following relation, which is not a solution of the one- dimensional wave equation?

Q: Out of the following relation, which is not a solution of the one- dimensional wave equation?

The one-dimensional wave equation is given by: \[ \frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2} \] To determine which of the given options is not a solution to this equation, we need to check each option by substituting it into the wave equation and verifying whether it satisfies the equation. **Option 1: $ y = x^2 + v^2 t^2 $** - Calculate the second partial derivatives: \[ \frac{\partial^2 y}{\partial t^2} = 2v^2 \] \[ \frac{\partial^2 y}{\partial x^2} = 2 \] - Substitute into the wave equation: \[ 2v^2 \neq v^2 \cdot 2 \] This does not satisfy the wave equation. **Option 2: $ y = x^2 - v^2 t^2 $** - Calculate the second partial derivatives: \[ \frac{\partial^2 y}{\partial t^2} = -2v^2 \] \[ \frac{\partial^2 y}{\partial x^2} = 2 \] - Substitute into the wave equation: \[ -2v^2 \neq v^2 \cdot 2 \] This does not satisfy the wave equation, confirming that Option 2 is not a solution. **Option 3: $ y = (x + vt)^2 $** - Calculate the second partial derivatives: \[ \frac{\partial^2 y}{\partial t^2} = 2v^2 \] \[ \frac{\partial^2 y}{\partial x^2} = 2 \] - Substitute into the wave equation: \[ 2v^2 = v^2 \cdot 2 \] This satisfies the wave equation. **Option 4: $ y = (x - vt)^2 $** - Calculate the second partial derivatives: \[ \frac{\partial^2 y}{\partial t^2} = 2v^2 \] \[ \frac{\partial^2 y}{\partial x^2} = 2 \] - Substitute into the wave equation: \[ 2v^2 = v^2 \cdot 2 \] This satisfies the wave equation. In conclusion, Option 2, $ y = x^2 - v^2 t^2 $, does not satisfy the one-dimensional wave equation, as the second derivatives do not equate as required by the wave equation. Therefore, Option 2 is correctly identified as not being a solution to the wave equation.

Question

Accepted Answer

The one-dimensional wave equation is given by:

$$
\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}
$$

To determine which of the given options is not a solution to this equation, we need to check each option by substituting it into the wave equation and verifying whether it satisfies the equation.

**Option 1: $ y = x^2 + v^2 t^2 $**

- Calculate the second partial derivatives:
  $$
  \frac{\partial^2 y}{\partial t^2} = 2v^2
  $$
  $$
  \frac{\partial^2 y}{\partial x^2} = 2
  $$
- Substitute into the wave equation:
  $$
  2v^2 \neq v^2 \cdot 2
  $$
  This does not satisfy the wave equation.

**Option 2: $ y = x^2 - v^2 t^2 $**

- Calculate the second partial derivatives:
  $$
  \frac{\partial^2 y}{\partial t^2} = -2v^2
  $$
  $$
  \frac{\partial^2 y}{\partial x^2} = 2
  $$
- Substitute into the wave equation:
  $$
  -2v^2 \neq v^2 \cdot 2
  $$
  This does not satisfy the wave equation, confirming that Option 2 is not a solution.

**Option 3: $ y = (x + vt)^2 $**

- Calculate the second partial derivatives:
  $$
  \frac{\partial^2 y}{\partial t^2} = 2v^2
  $$
  $$
  \frac{\partial^2 y}{\partial x^2} = 2
  $$
- Substitute into the wave equation:
  $$
  2v^2 = v^2 \cdot 2
  $$
  This satisfies the wave equation.

**Option 4: $ y = (x - vt)^2 $**

- Calculate the second partial derivatives:
  $$
  \frac{\partial^2 y}{\partial t^2} = 2v^2
  $$
  $$
  \frac{\partial^2 y}{\partial x^2} = 2
  $$
- Substitute into the wave equation:
  $$
  2v^2 = v^2 \cdot 2
  $$
  This satisfies the wave equation.

In conclusion, Option 2, $ y = x^2 - v^2 t^2 $, does not satisfy the one-dimensional wave equation, as the second derivatives do not equate as required by the wave equation. Therefore, Option 2 is correctly identified as not being a solution to the wave equation.