Question

Show that the function

i) u( x, y) = tan −1(y / x) is a solution of the two-dimensional Laplace’s equation.

ii)  u(x , t ) = e−6t cos 2x is a solution of the one-dimensional heat equation.


 

27 Jan 2018
Answer :
Word Count : 584
Let's go through both parts of the problem step by step. --- ### Part i) Showing that \( u(x, y) = \tan^{-1}\left(\frac{y}{x}\right) \) is a solution of the two-dimensional Laplace’s equation. The two-dimensional Laplace's equation is given by: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \] We are given the function \( u(x, y) = \tan^{-1}\left(\frac{y}{x}\right) \). #### Step 1: Compute the partial derivatives of \( u(x, y) \). First Partial Derivative with respect to \( x \): \[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \left( \tan^{-1}\left(\frac{y}{x}\right) \right) \] Using the chain rule: \[ \frac{\partial u}{\partial x} = \frac{-y}{x^2 + y^2} \] Second Partial Derivative with respect to \( x \): \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{-y}{x^2 + y^2} \right) \] Using the quotient rule: \[ \frac{\partial^2 u}{\partial x^2} = \frac{2xy^2 - 2x^3}{(x^2 +
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