Discrepancy in solutions of differential equation?











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The differential equation at hand is this :



$$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
And what I have tried is this :
$$
int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$

and
$$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$



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










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    up vote
    1
    down vote

    favorite












    The differential equation at hand is this :



    $$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
    And what I have tried is this :
    $$
    int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$

    and
    $$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$



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










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      The differential equation at hand is this :



      $$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
      And what I have tried is this :
      $$
      int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$

      and
      $$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$



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










      share|cite|improve this question















      The differential equation at hand is this :



      $$ frac{text{d}psi}{text{d}x}+2tanh(x),psileft(xright)=0 $$
      And what I have tried is this :
      $$
      int_{}^{} frac{text{d}psi}{psi}=-2int_{}^{} tanh(x),dx$$

      and
      $$ lnpsi left(xright)=-2cosh^{-2}left(xright)+C $$



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







      differential-equations hyperbolic-functions






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      edited Nov 14 at 11:19









      Robert Z

      90k1056128




      90k1056128










      asked Nov 14 at 11:00









      Jalaj Chaturvedi

      787




      787






















          1 Answer
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          up vote
          1
          down vote



          accepted










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






          share|cite|improve this answer























          • Oh! Thanks, I didn't thought about that.
            – Jalaj Chaturvedi
            Nov 14 at 11:12










          • @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
            – Robert Z
            Nov 14 at 11:32











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          1 Answer
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          active

          oldest

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          1 Answer
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          active

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          active

          oldest

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          active

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          up vote
          1
          down vote



          accepted










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






          share|cite|improve this answer























          • Oh! Thanks, I didn't thought about that.
            – Jalaj Chaturvedi
            Nov 14 at 11:12










          • @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
            – Robert Z
            Nov 14 at 11:32















          up vote
          1
          down vote



          accepted










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






          share|cite|improve this answer























          • Oh! Thanks, I didn't thought about that.
            – Jalaj Chaturvedi
            Nov 14 at 11:12










          • @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
            – Robert Z
            Nov 14 at 11:32













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






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






          share|cite|improve this answer














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







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 14 at 11:17

























          answered Nov 14 at 11:04









          Robert Z

          90k1056128




          90k1056128












          • Oh! Thanks, I didn't thought about that.
            – Jalaj Chaturvedi
            Nov 14 at 11:12










          • @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
            – Robert Z
            Nov 14 at 11:32


















          • Oh! Thanks, I didn't thought about that.
            – Jalaj Chaturvedi
            Nov 14 at 11:12










          • @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
            – Robert Z
            Nov 14 at 11:32
















          Oh! Thanks, I didn't thought about that.
          – Jalaj Chaturvedi
          Nov 14 at 11:12




          Oh! Thanks, I didn't thought about that.
          – Jalaj Chaturvedi
          Nov 14 at 11:12












          @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
          – Robert Z
          Nov 14 at 11:32




          @JalajChaturvedi Well done. BTW, if you are new here, please take a few minutes for a tour: math.stackexchange.com/tour
          – Robert Z
          Nov 14 at 11:32


















           

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