I entered a command incorrectly as follows:
DSolve[{y'[x]=y[x]},y[x],x]
I am now experiencing:
DSolve[{y'[x] == y[x]}, y[x], x]
During evaluation of In[26]:= DSolve::deqn: Equation or list of equations expected instead of True in the first argument {True}. >>
(* DSolve[{True},y(x),x] *)
How do I recover from this error. I've tried Clear[y'[x]]. That didn't work.
It is for situations like this that Unset exists. :-)
After the mistaken Set operation:
y'[x] = "oh dear";
y'[x]
"oh dear"
Merely use:
y'[x] =.
The definition is cleared:
y'[x]
y'[x]
Please see halirutan's answer for an explanation of why ClearAll[y] does not work here.
Very tricky mistake because hard to track down. The problem is that y'[x] parses as
Derivative[1][y][x]
Therefore, your assignment is not to the symbol y but to the symbol Derivative and since you have multiple call like f[][] it goes into its SubValues:
SubValues[Derivative]
(* {HoldPattern[Derivative[1][y][x]] :> y[x]} *)
Therefore, evaluate
SubValues[Derivative] = {};
and the sun shines again
DSolve[{y'[x] == y[x]}, y[x], x]
(* {{y[x] -> E^x C[1]}} *)
Deritative this non-specific clearing will not be appropriate. In version 7 for example there are definitions for InverseLaplaceTransform by default.
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– Mr.Wizard♦
Jan 13 '14 at 8:37
Unset is the way to go IMO
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– halirutan♦
Jan 13 '14 at 17:07
y'[x] is not the issue as one can see with HoldPattern[y'[x]] = "bad"; rather y'[x] parses as Derivative[1][y][x], as seen with HoldForm @ FullForm[y'[x]]. I am editing this answer accordingly.
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– Mr.Wizard♦
May 1 '16 at 17:39
Remove[y]will do the trick. $\endgroup$ – ciao Jan 13 '14 at 6:25Removehere! It will break any definitions that referencey, and it will alter even localized appearances of the Symbol, though the definitions may still work. Try e.g.fn[y_] := Sin[y]; Remove[y]; Definition[fn]$\endgroup$ – Mr.Wizard♦ Jan 13 '14 at 8:34