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# Solution of differential equations with boundary conditions contain "Limit"

Posted 10 years ago
 Dear Friends, I am going to solve the following differential equation: $y''[x] = {k^2}y[x]$ with boundary conditions is $y'[0] - hy[0] = 0$ and $\mathop {Lim}\limits_{x \to \infty } [\left| {y[x]} \right| < \infty$ I tried to solve it in Mathematica by DSolve[{y''[x] == {k^2}y[x],y'[0] - hy[0] == 0,Limit[Abs[y[x]], x -> \[Infinity] < \[Infinity]]},y[x],x]  But I get the following error: Equation or list of equations expected instead of Limit[Abs[y[x]],x->False] in the first argument. I can use the boundary conditions contain "Limit" how to correct? I would highly appreciate it if somebody can help me. Best regards.
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Posted 10 years ago
 This is what I expected!! I really appreciate your help with my problem! Thank you so much!!
Posted 10 years ago
 Thank you very much for helping me! But I still exist two problems: The boundary conditions of the problem is: $\mathop {Lim}\limits_{x \to \infty } [\left| {y[x]} \right| < \infty$. (y [x] is always finite). I can not use Plot command for $y[x]$ because the answer is still one constant C. Could I eliminate constant C by the second condition? how can i do that?Please help me one more!! I sincerely thank you very much!
Posted 10 years ago
 For a linear DE with linear BCs, there is always an arbitrary multiplicative constant in the solution. The constant is often determined by a normalization condition on f(x).
Posted 10 years ago
 Well, if C[1] is not equal zero, then the solution will diverge. So if you want the limit to be finite, i.e. $<\infty$, then the only thing that works is C[1]=0. So using that condition of yours, the only choice for the constant that works seems to be C[1]=0.Cheers,M.
Posted 10 years ago
 Syntax errors. Try following:In[1]:= DSolve[{y''[x] == k^2 y[x], y'[0] - h y[0] == 0}, y[x], x]Out[1]= {{y[x] -> ( E^(-k x) (-h + E^(2 k x) h + k + E^(2 k x) k) C[1])/(h + k)}}Clearly then Lim x->Infinity, gives y[x]->Infinity
Posted 10 years ago
 Unless C[1]=0, in which case the solution is constant zero, which appears to be a valid solution. M.
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