The differential equation at hand is this :

$$ \frac{\text{d}\psi}{\text{d}x}+2\tanh(x)\,\psi\left(x\right)=0\ $$ And what I have tried is this : $$ \int_{}^{} \frac{\text{d}\psi}{\psi}=-2\int_{}^{} \tanh(x)\,dx$$ and $$ \ln\psi \left(x\right)=-2\cosh^{-2}\left(x\right)+C\ $$

And the solution of this elementary problem comes out to be : $$ \psi\left(x\right)=Ae^{-2\cosh^{-2}\left(x\right)}$$ But clearly, $$ \psi\left(x\right)\ = \cosh^{-2}\left(x\right)\ $$ is a solution. But why can't I find it through integration?


1 Answer 1


Actually, we have that $$\int\tanh(x) dx=\int\frac{d(\cosh(x))}{\cosh(x)}=\ln(\cosh(x))+c$$ and
$$D(\tanh(x))=\frac{1}{\cosh^2(x)}.$$ So you confused the derivative with the integral...

  • $\begingroup$ Oh! Thanks, I didn't thought about that. $\endgroup$ Nov 14, 2018 at 11:12
  • $\begingroup$ @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour $\endgroup$
    – Robert Z
    Nov 14, 2018 at 11:32

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