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U(x,z)=X(x)Z(z)=an sin nπx L sinh nπz L,n=1,2,3,··(29) where an=c2 c4 is an arbitrary constant Since the PDE (1) is linear, we may superimpose solutions to obtain the form u(x,z)= X∞ n=1 an sin nπx L sinh nπz L (30) We still have to satisfy the BC (4), u(x,L)=1, which gives (from (30)) 1= X∞ n=1 an sin nπx L sinh nπ (31) EqPage 5 x y z P x y z C P 316) Simplify the following expressions, and implement them with twolevel NAND gate circuits a) AB′ ABD ABD′ A′C′D′ A′BC′Z b a f(x)g(x)dx = Z b a f(x)dx Z b a g(x)dx True This is one of the properties of definite integrals (b) If f and g are continuous on a,b, then Z b a f(x)g(x)dx = Z b a f(x)dxZ b a g(x)dx Oooh this is bad on so many levels! O ÿiauyu O A3 U O O E Ino Oo Uui 5 ˆ S C G µ µ Noyw µo C J Xeithd µz C Ssth O F 4 Zºmyª Y Cioypa O Id 3 Un X Acaei2i Stock Photo Alamy X zx x c c

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