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I have a problem in the form:

$-y''(x)+V(x)y(x)=0 $

with the initial conditions: $\quad y'(0)=k \ y(0), \quad k=cte$

The $V(x)$ is a model's parameter, for example $V(x)=\frac{6}{\left(x+2\right)^2}$.

In order to resolve it numerically for a given input $V(x)$, i try to write:

k=-50;
s = NDSolve[{-y''[x] + v[x]*y[x] == 0, y'[0] == k*y[0]}, y, {x,0, 100}]

As NDSolve requires two initial conditions the message shows:

"The number of constraints (1) (initial conditions) is not equal to the total differential order of the system plus the number of discrete variables (2). >>"

How to use NDsolve with vinculated initial conditions?

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closed as off-topic by MarcoB, user9660, Edmund, Alexey Popkov, Öskå Jun 15 '16 at 20:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Community, Edmund, Alexey Popkov, Öskå
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This is the problem, i cannot impose a specifics values on the y'(0) or y(0) . The v[x] depends of a model and is numeracally envaluated. $\endgroup$ – Melina A.H Jun 2 '16 at 6:41
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    $\begingroup$ maybe s = ParametricNDSolve[{-y''[x] + v[x]*y[x] == 0, y'[0] == k*a, y[0] == a}, y, {x, 0, 100}, {a}]? $\endgroup$ – kglr Jun 2 '16 at 6:49