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 particular equation, we need to substitute the function into the equation and see if it satisfies the equation. Let's start with the first part:
(i) To show that u = e^x * sin(y) is a solution of Laplace's equation, we need to substitute it into the Laplace's equation and check if it holds for all values of x and y.
Laplace's equation is given by ∇^2u = 0, where ∇^2 is the Laplacian operator. In two dimensions, the Laplacian operator is ∇^2 = ∂^2/∂x^2 + ∂^2/∂y^2.
Now let's substitute u = e^x * sin(y) into the Laplace's equation:
∇^2u = (∂^2u/∂x^2) + (∂^2u/∂y^2)
= (∂^2(e^x * sin(y))/∂x^2) + (∂^2(e^x * sin(y))/∂y^2)
We can calculate the second derivatives separately:
(∂^2(e^x * sin(y))/∂x^2) = (e^x * sin(y))" = e^x * sin(y)
(∂^2(e^x * sin(y))/∂y^2) = (e^x * sin(y))'' = e^x * sin(y)
Substituting these derivatives back into the equation, we have:
∇^2u = e^x * sin(y) + e^x * sin(y)
= 2 * e^x * sin(y)
Since 2 * e^x * sin(y) is not equal to zero for all values of x and y, we conclude that u = e^x * sin(y) is not a solution of Laplace's equation (∇^2u = 0).
(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 if it holds for all values of x and t.
The wave equation is given by ∂^2u/∂t^2 = c^2 * (∂^2u/∂x^2), where c is the wave speed.
Now let's substitute u = x^2 + t^2 into the wave equation:
∂^2u/∂t^2 = (∂^2(x^2 + t^2)/∂t^2) = 2
∂^2u/∂x^2 = (∂^2(x^2 + t^2)/∂x^2) = 2
Substituting these derivatives back into the equation, we have:
2 = c^2 * 2
For the equation to hold for all values of x and t, c^2 must be equal to 1. Hence, the wave equation is satisfied when c^2 = 1.
Therefore, u = x^2 + t^2 is a solution of the wave equation (∂^2u/∂t^2 = (∂^2u/∂x^2) with c^2 = 1).