Show that: (I) u = e^xsiny is a solution of Laplace's equation.

(ii) u = x^2+t^2 is a solution of the wave equation.

To show that a function is a solution of an equation, we need to substitute the function into the equation and verify that it satisfies the equation.

(i) To show that u = e^xsin(y) is a solution of Laplace's equation, we need to substitute u into the equation ∇^2u = 0 and show that it satisfies the equation.

Laplace's equation in two dimensions is given by:

∂^2u/∂x^2 + ∂^2u/∂y^2 = 0

Let's calculate the second derivatives of u with respect to x and y:

∂^2u/∂x^2 = ∂/∂x(e^xsin(y)) = e^xsin(y)

∂^2u/∂y^2 = ∂/∂y(e^xsin(y)) = e^xcos(y)

Substituting these derivatives back into Laplace's equation, we have:

e^xsin(y) + e^xcos(y) = 0

Since this equation holds true for all values of x and y, we can conclude that u = e^xsin(y) is a solution of Laplace's equation.

(ii) To show that u = x^2 + t^2 is a solution of the wave equation, we need to substitute u into the equation ∂^2u/∂t^2 - c^2∇^2u = 0 and show that it satisfies the equation.

The wave equation in three dimensions is given by:

∂^2u/∂t^2 - c^2(∂^2u/∂x^2 + ∂^2u/∂y^2 + ∂^2u/∂z^2) = 0

In this case, we have u = x^2 + t^2. Let's calculate the second derivatives with respect to t, x, y, and z:

∂^2u/∂t^2 = ∂/∂t(∂u/∂t) = ∂/∂t(2t) = 2

∂^2u/∂x^2 = ∂/∂x(∂u/∂x) = ∂/∂x(2x) = 2

∂^2u/∂y^2 = ∂/∂y(∂u/∂y) = ∂/∂y(0) = 0

∂^2u/∂z^2 = ∂/∂z(∂u/∂z) = ∂/∂z(0) = 0

Substituting these derivatives back into the wave equation, we have:

2 - c^2(2 + 0 + 0) = 2 - 2c^2 = 0

This equation holds true if 2 - 2c^2 = 0, or equivalently c = ± √(2). Therefore, for a specific value of c (in this case, c = ± √(2)), u = x^2 + t^2 is a solution of the wave equation.