Question

show that the function
1) u(x,y)=tan^-1(y/x) is a solution of the two dimensional laplaces equation.
2) u(x,t)=e^-(6t) cos2x is a solution of the one dimensional heat equation.

Answer 1

Answer on Question #73546 – Math – Differential Equations

Question

show that the function

1) $u(x,y) = \tan^{\wedge} - 1(y/x)$ is a solution of the two dimensional Laplace’s equation.

2) $u(x,t) = e^{\wedge} - (6t)\cos 2x$ is a solution of the one dimensional heat equation.

Solution

1)

$$\frac{\partial u}{\partial x} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2}, \quad \frac{\partial u}{\partial y} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2}.$$ $$\frac{\partial^2 u}{\partial x^2} = -y \cdot \left(-\frac{1}{(x^2 + y^2)^2}\right) \cdot 2x = \frac{2xy}{(x^2 + y^2)^2},$$ $$\frac{\partial^2 u}{\partial y^2} = x \cdot \left(-\frac{1}{(x^2 + y^2)^2}\right) \cdot 2y = -\frac{2xy}{(x^2 + y^2)^2},$$ $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} - \frac{2xy}{(x^2 + y^2)^2} = 0.$$

2)

$$\frac{\partial u}{\partial x} = -2e^{-6t}\sin 2x, \quad \frac{\partial u}{\partial t} = -6e^{-6t}\cos 2x.$$ $$\frac{\partial^2 u}{\partial x^2} = -4e^{-6t}\cos 2x.$$ $$\frac{3}{2}\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t} \quad \text{or} \quad \frac{\partial u}{\partial t} - k\frac{\partial^2 u}{\partial x^2} = 0 \quad \text{where} \quad k = \frac{3}{2}.$$

Answer provided by https://www.AssignmentExpert.com