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 given function is a solution of a certain differential equation, we need to substitute the function into the differential 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 it into the Laplace's equation and check whether it holds.

Laplace's equation is given by:

∇^2u = ∂^2u/∂x^2 + ∂^2u/∂y^2 = 0,

where ∇^2 denotes the Laplacian operator.

Let's evaluate the second derivatives with respect to x and y:

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

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

Now, substitute these results into the Laplace's equation:

∂^2u/∂x^2 + ∂^2u/∂y^2 = e^(xsin(y)) * cos(y) * sin(y) + e^(xsin(y)) * x * cos(y) * sin(y)

Since we are looking for a solution where ∇^2u = 0, the above expression should simplify to 0.

Notice that both terms have a common factor of e^(xsin(y)) * cos(y) * sin(y). By factoring out this common factor, we get:

e^(xsin(y)) * cos(y) * sin(y) * (1 + x) = 0

For this equation to hold, either e^(xsin(y)) * cos(y) * sin(y) = 0 or (1 + x) = 0.

Since (1 + x) = 0 is not true for all x, we focus on the first case:

e^(xsin(y)) * cos(y) * sin(y) = 0

To satisfy this equation, we must have e^(xsin(y)) = 0 or cos(y) * sin(y) = 0.

However, e^(xsin(y)) cannot be zero for any values of x and y because the exponential function e^z is always positive for any real value of z. Therefore, e^(xsin(y)) = 0 does not hold.

We are left with cos(y) * sin(y) = 0.

This equation holds for y = nπ, where n is any integer. However, we are interested in solutions that are not identically zero.

Hence, we conclude that u = e^(xsin(y)) is not 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 it into the wave equation and check whether it holds.

The wave equation is given by:

∂^2u/∂t^2 - c^2∇^2u = 0,

where c is the wave speed, and ∇^2 denotes the Laplacian operator.

First, let's evaluate the second derivative with respect to t:

∂^2u/∂t^2 = ∂^2/∂t^2(x^2 + t^2)
= 2

Now, evaluate the Laplacian of u:

∇^2u = ∂^2u/∂x^2 + ∂^2u/∂y^2 + ∂^2u/∂z^2

Since u is given as a function of x and t only, there are no y- or z- derivatives involved. Therefore, this simplifies to:

∇^2u = ∂^2u/∂x^2

Now, evaluate the second derivative with respect to x:

∂^2u/∂x^2 = ∂^2/∂x^2(x^2 + t^2)
= 2

Now, substitute the results into the wave equation:

∂^2u/∂t^2 - c^2∇^2u = 2 - c^2 * 2

This expression can be simplified to:

2 - 2c^2

For this equation to hold, 2 - 2c^2 = 0.

Simplifying the equation further, we find:

c^2 = 1

Therefore, the wave equation holds for any value of c^2 = 1.

Hence, we can conclude that u = x^2 + t^2 is indeed a solution of the wave equation.