Question

(2 points) is typed as lambda, a as alpha. The PDE yº au au ar ay is separable, so we look for solutions of the form u(x, t) = X(2)Y(y). When solving DE in X and Y use the constants a and b for X and c for Y. The PDE can be rewritten using this solution as (placing constants in the DE for Y) into X"/X = (1/(k^2))(y^5)(Y'/Y) -2 Note: Use the prime notation for derivatives, so the derivative of X is written as X'. DO NOT use X'(2) Since these differential equations are independent of each other, they can be separated DE in X: X"+lambdax =0 DE in Y: lambdak2Y+y^5Y' = 0 = Now we solve the separate separated ODEs for the different cases in 1. In each case the general solution in X is written with constants a and b and the general solution in Y is written with constants c and d. Write the functions alphabetically, so that if the solutions involve cos and sin, your answer would be acos(x) + bsin(x). Case 1: = 0 X(X) = a+bx Y(y) с DE in Y If + 0, the differential equation in Y is first order, linear, and more importantly separable. We separate the two sides as Y/Y = -((k^2)/(y^5) lambda Integrating both sides with respect to y (placing the constant of integration c in the right hand side) we get In(absY) = KA2lambda/(4y 4)+C Solving for Y, using the funny algebra of constant where e = cis just another constant we get Y = ce^(k^2lambda/(4y^4)) For 1 + 0 we get a Sturm-Louiville problem in X which we need to handle two more cases Case 2: X=-a? X(X) = a*cosh(alphax)+b*sinh(alphax) Case 3: X=a2 X(x) = a*cos(alphax)+b*sin(alphax)
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Answer 1