1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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...

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.