Suppose that $ b_1 = const .f_1(x) \left( b_2 + \frac{a} {f_2(x)} \right) $ becomes $ b_1 = const. b_2 $ because $f_1(x) \rightarrow 1$ and $f_2(x) \rightarrow \infty$ when $x \rightarrow \infty$. Do you agree that it allows us to write $ const.b_2 = const .f_1(x) \left( b_2 + \frac{a} {f_2(x)} \right) ?$
Is the following substitution legitimate?
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algebra-precalculus
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0@ganzewoort I'm sorry but I don't know exactly what$a$"boundary value problem" is since I've never studied differential equations but hopefully someone else can help you on that :) – 2011-09-23
1 Answers
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The substitution is not directly valid because:
b1=Constant.b2
is not an absolute equality. It is a constrained equality by the constraint x --> infinity, so you can't just use it without that constraint.
You can see an example if you use this specific case: f1(x) = (1+1/x) and f2(x)=x
Hope this helps.
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0Well, I guess the case is not the same. To my limited knowledge, in boundary problems of Diff. Equations, the result is a function that obeys the boundary conditions (constraints). That result is not the generalized result, however, it obeys the original condition under the given constraints only. A good example may be the one in http://en.wikipedia.org/wiki/Boundary_value_problem in this case, y(x)=2 Sin(x) is not the general result but a special one satisfying y(0)=0 and y(pi/2)=2 – 2011-09-23