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# Important Instructions: (1) λ is typed as lambda. (2) Use hyperbolic trig functions cosh(x) and sinh(x)...

Important Instructions: (1) λ is typed as lambda. (2) Use hyperbolic trig functions cosh(x) and sinh(x) instead of ex and e−x. (3) Write the functions alphabetically, so that if the solutions involve cos and sin, your answer would be Acos(x)+Bsin(x). (4) For polynomials use arbitrary constants in alphabetical order starting with highest power of x, for example, Ax2+Bx. (5) Write differential equations with leading term positive, so X′′−2X=0 rather than −X′′+2X=0. (6) Finally you need to simplify arbitrary constants. For example if A, B and C are arbitrary constants then AC, BC could be simplified to A, B since C in this case would be redundant. The PDE k2∂2u∂x2=y2∂u∂y is separable, so we look for solutions of the form u(x,t)=X(x)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 (place constant k in the DE for Y) into X''/X = (1/k^2)(y^2)(Y'/Y) = −λ Note: Use the prime notation for derivatives, so the derivative of X and Y can be written as X′ and Y′. Do NOT use X′(x) Since these differential equations are independent of each other, they can be separated DE in X: X''+lambdaX =0 DE in Y: =0 (your answer should be simplified so there are no fractions) Now we solve the separated ODEs for the different cases in λ. 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. Case 1: λ=0 X(x)= Ax+B Y(y)= C DE in Y If λ≠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^2))lambda Integrating both sides with respect to y (placing the constant of integration C in the right hand side) we get ln(absY) = Solving for Y, using the funny algebra of constants where eC=C is just another constant we get Y= For λ≠0 we get a Sturm-Louiville problem in X which we need to handle two more cases Case 2: λ=−a2 X(x)= Case 3: λ=a2 X(x)= Final Producet Solution (make sure to remove any redundant arbitrary constants) Case 1: λ=0 u= Ax+B Case 2: λ=−a2 u= Case 3: λ=a2 u=

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