By Milovanov

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9) On the other hand, the boundary condition at x = b, Eq. 10) Since (Xo(b) and (X1(b) are now known quantities, Eqs. 12) As a result, Eq. 1) is tranformed into an initial value problem, since now Eq. 1) can be integrated backward from x = b by using the initial condition given by Eqs. 12). Another approach is to integrate Eq. 11) as the initial condition. 2 will now be applied to three examples. To demonstrate its accuracy, solutions calculated by the method of chasing are compared with the exact solutions.

9) subject to the boundary conditions: yeO) = 0, dy(O) - - =0 dx ' y(l) =0 Comparison with Eq. 1) shows that fl(x) = 0, f2(x) = -7, f3(x) = 6, rex) = 6 22 2. Integration of Eqs. 8007 which can be substituted into Eq. 6871 The solution of the given differential equation can therefore be calculated by using Eq. 3) since now YI(X), h(x), and the constant J-t are known. Because of its simplicity, no details will be given. 1 Three-Point Third-Order Differential Equations It should be noted that the form of Eqs.

5 Schematic diagram of the fin. Simplifying Eq. , x = 0: T= T, On the other surface, x = L, the plate is exposed to air and the only mode of heat transfer is by convection. Equating the conduction out of the solid to the convection into the fluid, we get x= L: -kdT(L)/dx = ii[T(L) - Too] If the following dimensionless quantities are introduced, x = x/ L, Eq. 29) 40 3. Method of Chasing where f3 = m2L 2 , N bi = fiLl k Eq. 27) appears to be simple to solve. However, if we compare Eqs. 29) with Eqs.