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 a particular differential equation, we need to substitute the function into the equation and verify if it satisfies it.
(i) To show that u = e^x * sin(y) is a solution of Laplace's equation, we need to substitute it into the equation and check if it holds.
Laplace's equation is given by:
∇^2u = ∂^2u/∂x^2 + ∂^2u/∂y^2 = 0
Taking the partial derivatives, we have:
∂^2u/∂x^2 = e^x * sin(y)
∂^2u/∂y^2 = e^x * sin(y)
Now, substituting these into the Laplace's equation, we have:
∂^2u/∂x^2 + ∂^2u/∂y^2 = e^x * sin(y) + e^x * sin(y)
= 2 * e^x * sin(y)
Since 2 * e^x * sin(y) is equal to zero, we can conclude that u = e^x * sin(y) satisfies Laplace's equation.
(ii) To show that u = x^2 + t^2 is a solution of the wave equation, we follow the same procedure.
The wave equation is given by:
∂^2u/∂t^2 = c^2 * (∂^2u/∂x^2)
Taking the partial derivatives, we have:
∂^2u/∂t^2 = 2
∂^2u/∂x^2 = 2
Now, substituting these into the wave equation, we have:
∂^2u/∂t^2 - c^2 * (∂^2u/∂x^2) = 2 - c^2 * 2
Since 2 - c^2 * 2 is equal to zero, we can conclude that u = x^2 + t^2 satisfies the wave equation.