Solution of the differential equation $ddot{x}=e^{-x}$












2












$begingroup$


I would like to know how to get the solution of the following differential equation:



$$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$



A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.






ordinary-differential-equations







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$endgroup$

















    2












    $begingroup$


    I would like to know how to get the solution of the following differential equation:



    $$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$



    A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.






    ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I would like to know how to get the solution of the following differential equation:



      $$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$



      A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.






      ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      I would like to know how to get the solution of the following differential equation:



      $$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$



      A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.






      ordinary-differential-equations




      ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 16:15









      Élio PereiraÉlio Pereira

      402514




      402514






















          1 Answer
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          $begingroup$

          We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.



          $$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.



          Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$






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          • 2




            $begingroup$
            You can also use $u=e^{x/2}$. Lots of possibilities!
            $endgroup$
            – Dylan
            Jan 7 at 16:47










          • $begingroup$
            Thank you very much for the help!
            $endgroup$
            – Élio Pereira
            Jan 7 at 16:49











          Your Answer





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          1 Answer
          1






          active

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          active

          oldest

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          10












          $begingroup$

          We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.



          $$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.



          Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            You can also use $u=e^{x/2}$. Lots of possibilities!
            $endgroup$
            – Dylan
            Jan 7 at 16:47










          • $begingroup$
            Thank you very much for the help!
            $endgroup$
            – Élio Pereira
            Jan 7 at 16:49
















          10












          $begingroup$

          We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.



          $$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.



          Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            You can also use $u=e^{x/2}$. Lots of possibilities!
            $endgroup$
            – Dylan
            Jan 7 at 16:47










          • $begingroup$
            Thank you very much for the help!
            $endgroup$
            – Élio Pereira
            Jan 7 at 16:49














          10












          10








          10





          $begingroup$

          We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.



          $$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.



          Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$






          share|cite|improve this answer











          $endgroup$



          We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.



          $$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.



          Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 7 at 16:39

























          answered Jan 7 at 16:25









          John DoeJohn Doe

          11.1k11238




          11.1k11238








          • 2




            $begingroup$
            You can also use $u=e^{x/2}$. Lots of possibilities!
            $endgroup$
            – Dylan
            Jan 7 at 16:47










          • $begingroup$
            Thank you very much for the help!
            $endgroup$
            – Élio Pereira
            Jan 7 at 16:49














          • 2




            $begingroup$
            You can also use $u=e^{x/2}$. Lots of possibilities!
            $endgroup$
            – Dylan
            Jan 7 at 16:47










          • $begingroup$
            Thank you very much for the help!
            $endgroup$
            – Élio Pereira
            Jan 7 at 16:49








          2




          2




          $begingroup$
          You can also use $u=e^{x/2}$. Lots of possibilities!
          $endgroup$
          – Dylan
          Jan 7 at 16:47




          $begingroup$
          You can also use $u=e^{x/2}$. Lots of possibilities!
          $endgroup$
          – Dylan
          Jan 7 at 16:47












          $begingroup$
          Thank you very much for the help!
          $endgroup$
          – Élio Pereira
          Jan 7 at 16:49




          $begingroup$
          Thank you very much for the help!
          $endgroup$
          – Élio Pereira
          Jan 7 at 16:49


















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