Given the following two dimensional laplace partial...

Given the following two-dimensional Laplace partial differential equation (PDE): ∂²u/∂x² + ∂²u/∂y² = 0 Confined to a square 0 < x < 1 and 0 < y < 1 and with the following conditions: u(0,y) = 0, u(x,0) = 0, u(1,y) = 0, u(x,1) = 2x + 1 Using the method of separation of variables, demonstrate through the four main stages how to obtain the following solution to the PDE: u(x,y) = (2/(nπ)) sinh(nπ x) sin(nπ y) sinh(nπ(1-y)), where n is a positive integer. (10 marks) Hint: u(x,y) = X(x) Y(y) = Σ_{n=1}^∞ X_n Y_n

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00:01 This question we have to find out the solution of the given partial differential equation, okay, so this is first of all we have laplace equation uxx plus uyy equals to 0, i call this as equation 1 and x is xy, they are between 0 and 1, so using the separation of variable by separation of variable what we get, we take that u is equals to x of x times y of y is the solution correct, so from this what i am going to get, we put this value here from equation 1, u we know, so uxx comes out to be x double dash y plus y double dash x equals to 0, this will implies x double dash over x equals to minus y double dash over y is equals to let's say lambda, i call it as minus p square, this is equation number 2, so here we have taken lambda less than 0 which i call it as the separation constant and p is a number greater than 0, so from this what i will get from equation 2, if i consider x double dash over x is equals to lambda which is minus p square, so this implies x double dash plus p square x equals to 0, similarly for y also we get y double dash minus p square y equals to 0, this from this i get x is equals to c1 cosine of px plus c2 sine of px, from this we are getting y equals to c3 e raised to py plus c4 e raised to minus py, because if we consider the auxiliary equation, we get m square minus p square equals to 0, that means m comes out to be plus minus p, so we are getting this type of solution and here we get the imaginary, so we are getting this solution, correct.

02:48 Now we have the boundary conditions for x as follows, which is u of 0 y equals to 0, this will implies that x0 equals to 0, from this what we are going to get, here we are putting, so we get 0 equals to c1 times 1 plus c2 sine 0 is 0, this implies c1 value is 0 and the other condition is u1 y equals to 0, this again implies x1 equals to 0, we have to use again the same equation, we get c2 sine p equals to 0, from this we are getting sine p equals to sine of n pi, that means we got p as n pi or n value as 1, 2 and so on.

03:49 Now similarly we apply the boundary condition for y, what we are going to get for y if we apply, we have ux0 equals to 0, which implies y0 is 0, from the solution of y we get, if we put x equals to 0, we are getting c3, this is c3 plus c4 equals to 0, that means c3 is equals to minus c4, then we can write down x as c2 sine n pi x and my y now becomes c3 e raised to n pi y minus e raised to minus n pi y, this is nothing but c3 hyperbolic sine of n pi y, hence the solution u, which was of the form xy comes out to be cn sine n pi x times hyperbolic sine of n pi y, this will be the required solution where n varies from 1, 2, 3 and so on, this is a natural number...

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