I have an ODE described by
$$\frac{dk}{du} = \frac{\partial(ku-k^3)/\partial u}{V(u) - \partial(ku-k^3)/\partial k} $$
and I try to solve with
Dsolve[k'[u] == D[k*u - k^3, u]/(u - D[k*u - k^3,
k]), u, k]
But it tells me off. Why?
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2 Answers
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DSolve[k'[u] == D[k[u]*u - k[u]^3, u]/(u - D[k[u]*u - k[u]^3, k[u]]),
k[u], u]
This could work. The k need to be writen as k[u].
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Did you try that one?
D[k*u - k^3, u]/(u - D[k*u - k^3, k]);
DSolve[k'[u] == 1/(3 k[u]), k[u], u]
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$\begingroup$ This seems to be incorrect as k is a function of u, which will give you a completely different equation. See the other answer. $\endgroup$– DunlopApr 5, 2022 at 15:15
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1$\begingroup$ @Dunlop I think this is right, if $V(u)=u$ and the partial derivatives mean what they appear to mean in the TeX in the OP. There's some ambiguity about their meaning, but treating the dependent variable as a variable happens from time to time. Think $m x''(t) = -\partial U/\partial x$ where $U$ is the potential energy, for instance. $\endgroup$ Apr 5, 2022 at 15:49
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$\begingroup$ As Dunlop sad, I used the partial derivatives only in explicit variables. $\endgroup$ Apr 5, 2022 at 17:43
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$\begingroup$ I guess it then depends a bit on what he OP actually wanted which is somewhat ambiguous. $\endgroup$– DunlopApr 5, 2022 at 19:07
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$\begingroup$ Or even if they know what they want....I mean, what are the chances that the numerator should have been $d(ku-k^3)/du$, as in the other answer, but the denominator should be a partial derivative of the same function? $\endgroup$ Apr 5, 2022 at 19:49
kis a function ofuas it seems.uis a function of what? $\endgroup$Mmathat k is the independent variable and also k depends on u. $\endgroup$DSolve(capitalS) was misspelled in the OP. :) Did you useDsolveorDSolvewhen you ran your code? If you get error messages, please include them in the question. Tx. $\endgroup$