Temperature Distribution Finite-Difference Boundary-Value ExampleBar and heat transfer info:≔L10 length ≔TA40 temperature at left end≔k0.1convection/conduction constant≔TB200temperature at right end≔n20number of intervals≔Ta20ambient temperatureODE:＝+――dd2x2T⋅k⎛⎝−TaT⎞⎠0BC's:＝T((0))TA＝T((L))TBTheoretical solution (using methods of Differential Equations):＝−――dd2x2T⋅kT ⋅−k Tahomogeneous solution:―――→＝−s2k0,solve sexplicit‾‾k−‾‾k⎡⎢⎢⎣⎤⎥⎥⎦＝Th((x))+⋅C1e⋅‾kx⋅C2e⋅−‾kxparticular solution:≔Tp((x))Tageneral solution:＝＝T((x))+Th((x))Tp((x))++⋅C1e⋅‾kx⋅C2e⋅−‾kxTaapplying the BCs:＝＝T((0))TA++C1C2Ta＝＝T((L))TB++⋅C1e⋅‾kL⋅C2e⋅−‾kLTaNon-Commercial Use Onlyusing algebra:≔C1―――――――――−−TBTa⋅⎛⎝−TATa⎞⎠⋅−‾kL⎛⎜⎝−⋅‾kL ⋅−‾kL⎞⎟⎠=C17.597≔C2−−TATaC1=C212.403using MathCAD:Guess ValuesConstraintsSolver≔C17≔C210＝TA++C1C2Ta＝TB++⋅C1e⋅‾kL⋅C2e⋅−‾kLTa≔C1C2⎡⎢⎣⎤⎥⎦=⎛⎝,C1C2⎞⎠7.59712.403⎡⎢⎣⎤⎥⎦true analytical solution:≔Tt((x))++⋅C1⋅‾kx⋅C2⋅−‾kxTaMathCAD numerical solution:Guess ValuesConstraintsSolver＝+――dd2x2T((x))⋅k⎛⎝−TaT((x))⎞⎠0＝T((0))TA＝T((L))TB≔TM((,T((x))L))Non-Commercial Use OnlyODE (finite difference form):≔Δx―Lninterval size≔α+2⋅kΔx2≔β⋅⋅kΔx2Ta＝−+−T−i1⋅αTiT+i1β‥＝i1((−n1))Matrix form of finite difference equations:＝AT B≔i‥0((−n2))≔j‥0((−n2))≔m‥1((−n2))≔A,ij0≔A,iiα≔A,m−m1−1 ≔A,−m1m−1≔B0+βTA≔Bmβ≔B−n2+βTB=A2.025−100000−12.025−100000−12.025−100000−12.025−100000−12.025−100000−12.025−100000−12.025000000−1000000000000000000000⋱⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦=B40.50.50.50.50.50.50.50.50.50.50.50.5⋮⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦≔X⋅A−1B≔T0TA≔TnTB≔T+i1XiNon-Commercial Use Only≔i‥0n ≔xi⋅iΔx65809511012514015517018520035502152345678901 10TiTt⎛⎝xi⎞⎠TM⎛⎝xi⎞⎠xiNon-Commercial Use