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This question already has an answer here:

There is an equation for example:

eqn=D[c[x, t], t] == d D[c[x, t], x, x];

When I make a LaplaceTransform of it:

LaplaceTransform[eqn, t, s]

Then mathematica will impose the transform to both sides of the '=', and the result will be :

s*LaplaceTransform[c[x, t], t, s] - c[x, 0] == d*LaplaceTransform[Derivative[2, 0][c][x, t], t, s]

But when I make a FourierTransform it won't do that, and the output :

FourierTransform[Derivative[0, 1][c][x, t] == d*Derivative[2, 0][c][x, t], t, \[Omega]]

And I just wonder know why. What's the difference?

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marked as duplicate by xzczd differential-equations Nov 10 '18 at 12:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I think the only difference is that some developer told LaplaceTransform to do so while it came to nobody's mind to do the same for FourierTransform. So, no deep mathematics behind that.

If you like, you can add a the following definition to FourierTransform:

Unprotect[FourierTransform];
FourierTransform[eq_Equal, args__] := FourierTransform[#, args] & /@ eq;
Protect[FourierTransform];

Then you obtain

FourierTransform[Derivative[0, 1][c][x, t] == d*Derivative[2, 0][c][x, t], t, ω]

(-I)ωFourierTransform[c[x, t], t, ω] == FourierTransform[d*Derivative[2, 0][c][x, t], t, ω]

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