Solution to this Differential Equation needed

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I came up with this differential equation and I don't know how to solve it.



$$f''(x)=f(x)f'(x)$$



I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is



$$f(x)=sqrt2a tanleft(fracsqrt2a2 cdot (x+b)right),$$ where $a$ and $b$ are constants.



How does one get this solution?






calculus differential-equations







share|cite|improve this question



























    up vote
    2
    down vote

    favorite












    I came up with this differential equation and I don't know how to solve it.



    $$f''(x)=f(x)f'(x)$$



    I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is



    $$f(x)=sqrt2a tanleft(fracsqrt2a2 cdot (x+b)right),$$ where $a$ and $b$ are constants.



    How does one get this solution?






    calculus differential-equations







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I came up with this differential equation and I don't know how to solve it.



      $$f''(x)=f(x)f'(x)$$



      I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is



      $$f(x)=sqrt2a tanleft(fracsqrt2a2 cdot (x+b)right),$$ where $a$ and $b$ are constants.



      How does one get this solution?






      calculus differential-equations







      share|cite|improve this question















      I came up with this differential equation and I don't know how to solve it.



      $$f''(x)=f(x)f'(x)$$



      I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is



      $$f(x)=sqrt2a tanleft(fracsqrt2a2 cdot (x+b)right),$$ where $a$ and $b$ are constants.



      How does one get this solution?






      calculus differential-equations




      calculus differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      mwt

      534112




      534112










      asked 1 hour ago









      clathratus

      487




      487




















          2 Answers
          2






          active

          oldest

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          up vote
          4
          down vote













          $$y''=y'y implies y''=frac 12 (y^2)' $$
          After integration
          $$implies y'=frac 12 y^2+K$$
          $$y'=frac 12(y^2+2K)$$
          For $ K=0 implies -frac 1y=frac x2 +a$
          $$implies y(x)=-frac 2 x+c$$



          For K negative
          $$ int frac dyy^2-c^2=frac x 2+b$$
          $$ln (frac y-cy+c)=cx+a$$
          $$y(x)=-c frac ae^cx+1ae^cx-1$$



          For K positive, substitute $2K=c^2$



          $$ int frac dyy^2+c^2=frac 12int dx=frac x 2+b$$
          Substitute $z=y/c implies dz=dy/c$
          $$ int frac dzz^2+1=c(frac x 2+b)$$
          $$ arctan (z)=c(frac x 2+b)$$
          $$ y=ctan (c(frac x 2+b))$$
          Substitute $c/2=k$ and $bc=a$
          $$ y=2ktan (kx+a)$$



          $$y''=y'y$$
          Substitute
          $$y'= p implies y''=pp'$$
          $$p'=y implies p=frac 12 y^2 +K$$
          $$y'=frac 12 y^2+K$$
          This equation is separable






          share|cite|improve this answer






















          • I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation.
            – clathratus
            1 hour ago






          • 1




            just evaluate the integral @clathratus
            – Isham
            1 hour ago






          • 1




            I added some lines is it more clear now ? @clathratus
            – Isham
            54 mins ago






          • 1




            true @WillJagy I solved it for K positive I corrected my answer thanks a lot
            – Isham
            34 mins ago







          • 1




            @WillJagy Thanks a lot for your answer I have upvoted it
            – Isham
            12 mins ago

















          up vote
          1
          down vote













          as an example, in Isham's answer, if we choose $K$ so that
          $$ y' = frac12 (y^2-1), $$
          the first thing we notice is two stationary solutions, $y = 1$ and $y = -1$



          This is autonomous...



          For $y > 1 $ or $y < -1$ we get
          $$ y = frac-1tanh 2x $$
          which blow up in finite time, one in each direction. Also translates.



          For $-1 < y < 1$
          $$ y = - tanh 2x $$
          enter image description here






          share|cite|improve this answer


















          • 1




            @thanks a lot for your answer and the picture...that makes things more clear...
            – Isham
            4 mins ago










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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

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          oldest

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          active

          oldest

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          up vote
          4
          down vote













          $$y''=y'y implies y''=frac 12 (y^2)' $$
          After integration
          $$implies y'=frac 12 y^2+K$$
          $$y'=frac 12(y^2+2K)$$
          For $ K=0 implies -frac 1y=frac x2 +a$
          $$implies y(x)=-frac 2 x+c$$



          For K negative
          $$ int frac dyy^2-c^2=frac x 2+b$$
          $$ln (frac y-cy+c)=cx+a$$
          $$y(x)=-c frac ae^cx+1ae^cx-1$$



          For K positive, substitute $2K=c^2$



          $$ int frac dyy^2+c^2=frac 12int dx=frac x 2+b$$
          Substitute $z=y/c implies dz=dy/c$
          $$ int frac dzz^2+1=c(frac x 2+b)$$
          $$ arctan (z)=c(frac x 2+b)$$
          $$ y=ctan (c(frac x 2+b))$$
          Substitute $c/2=k$ and $bc=a$
          $$ y=2ktan (kx+a)$$



          $$y''=y'y$$
          Substitute
          $$y'= p implies y''=pp'$$
          $$p'=y implies p=frac 12 y^2 +K$$
          $$y'=frac 12 y^2+K$$
          This equation is separable






          share|cite|improve this answer






















          • I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation.
            – clathratus
            1 hour ago






          • 1




            just evaluate the integral @clathratus
            – Isham
            1 hour ago






          • 1




            I added some lines is it more clear now ? @clathratus
            – Isham
            54 mins ago






          • 1




            true @WillJagy I solved it for K positive I corrected my answer thanks a lot
            – Isham
            34 mins ago







          • 1




            @WillJagy Thanks a lot for your answer I have upvoted it
            – Isham
            12 mins ago














          up vote
          4
          down vote













          $$y''=y'y implies y''=frac 12 (y^2)' $$
          After integration
          $$implies y'=frac 12 y^2+K$$
          $$y'=frac 12(y^2+2K)$$
          For $ K=0 implies -frac 1y=frac x2 +a$
          $$implies y(x)=-frac 2 x+c$$



          For K negative
          $$ int frac dyy^2-c^2=frac x 2+b$$
          $$ln (frac y-cy+c)=cx+a$$
          $$y(x)=-c frac ae^cx+1ae^cx-1$$



          For K positive, substitute $2K=c^2$



          $$ int frac dyy^2+c^2=frac 12int dx=frac x 2+b$$
          Substitute $z=y/c implies dz=dy/c$
          $$ int frac dzz^2+1=c(frac x 2+b)$$
          $$ arctan (z)=c(frac x 2+b)$$
          $$ y=ctan (c(frac x 2+b))$$
          Substitute $c/2=k$ and $bc=a$
          $$ y=2ktan (kx+a)$$



          $$y''=y'y$$
          Substitute
          $$y'= p implies y''=pp'$$
          $$p'=y implies p=frac 12 y^2 +K$$
          $$y'=frac 12 y^2+K$$
          This equation is separable






          share|cite|improve this answer






















          • I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation.
            – clathratus
            1 hour ago






          • 1




            just evaluate the integral @clathratus
            – Isham
            1 hour ago






          • 1




            I added some lines is it more clear now ? @clathratus
            – Isham
            54 mins ago






          • 1




            true @WillJagy I solved it for K positive I corrected my answer thanks a lot
            – Isham
            34 mins ago







          • 1




            @WillJagy Thanks a lot for your answer I have upvoted it
            – Isham
            12 mins ago












          up vote
          4
          down vote










          up vote
          4
          down vote









          $$y''=y'y implies y''=frac 12 (y^2)' $$
          After integration
          $$implies y'=frac 12 y^2+K$$
          $$y'=frac 12(y^2+2K)$$
          For $ K=0 implies -frac 1y=frac x2 +a$
          $$implies y(x)=-frac 2 x+c$$



          For K negative
          $$ int frac dyy^2-c^2=frac x 2+b$$
          $$ln (frac y-cy+c)=cx+a$$
          $$y(x)=-c frac ae^cx+1ae^cx-1$$



          For K positive, substitute $2K=c^2$



          $$ int frac dyy^2+c^2=frac 12int dx=frac x 2+b$$
          Substitute $z=y/c implies dz=dy/c$
          $$ int frac dzz^2+1=c(frac x 2+b)$$
          $$ arctan (z)=c(frac x 2+b)$$
          $$ y=ctan (c(frac x 2+b))$$
          Substitute $c/2=k$ and $bc=a$
          $$ y=2ktan (kx+a)$$



          $$y''=y'y$$
          Substitute
          $$y'= p implies y''=pp'$$
          $$p'=y implies p=frac 12 y^2 +K$$
          $$y'=frac 12 y^2+K$$
          This equation is separable






          share|cite|improve this answer














          $$y''=y'y implies y''=frac 12 (y^2)' $$
          After integration
          $$implies y'=frac 12 y^2+K$$
          $$y'=frac 12(y^2+2K)$$
          For $ K=0 implies -frac 1y=frac x2 +a$
          $$implies y(x)=-frac 2 x+c$$



          For K negative
          $$ int frac dyy^2-c^2=frac x 2+b$$
          $$ln (frac y-cy+c)=cx+a$$
          $$y(x)=-c frac ae^cx+1ae^cx-1$$



          For K positive, substitute $2K=c^2$



          $$ int frac dyy^2+c^2=frac 12int dx=frac x 2+b$$
          Substitute $z=y/c implies dz=dy/c$
          $$ int frac dzz^2+1=c(frac x 2+b)$$
          $$ arctan (z)=c(frac x 2+b)$$
          $$ y=ctan (c(frac x 2+b))$$
          Substitute $c/2=k$ and $bc=a$
          $$ y=2ktan (kx+a)$$



          $$y''=y'y$$
          Substitute
          $$y'= p implies y''=pp'$$
          $$p'=y implies p=frac 12 y^2 +K$$
          $$y'=frac 12 y^2+K$$
          This equation is separable







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 6 mins ago

























          answered 1 hour ago









          Isham

          11.3k3929




          11.3k3929











          • I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation.
            – clathratus
            1 hour ago






          • 1




            just evaluate the integral @clathratus
            – Isham
            1 hour ago






          • 1




            I added some lines is it more clear now ? @clathratus
            – Isham
            54 mins ago






          • 1




            true @WillJagy I solved it for K positive I corrected my answer thanks a lot
            – Isham
            34 mins ago







          • 1




            @WillJagy Thanks a lot for your answer I have upvoted it
            – Isham
            12 mins ago
















          • I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation.
            – clathratus
            1 hour ago






          • 1




            just evaluate the integral @clathratus
            – Isham
            1 hour ago






          • 1




            I added some lines is it more clear now ? @clathratus
            – Isham
            54 mins ago






          • 1




            true @WillJagy I solved it for K positive I corrected my answer thanks a lot
            – Isham
            34 mins ago







          • 1




            @WillJagy Thanks a lot for your answer I have upvoted it
            – Isham
            12 mins ago















          I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation.
          – clathratus
          1 hour ago




          I got to that point by myself, but how do I get to the solution? I do not know the significance of a separate differential equation.
          – clathratus
          1 hour ago




          1




          1




          just evaluate the integral @clathratus
          – Isham
          1 hour ago




          just evaluate the integral @clathratus
          – Isham
          1 hour ago




          1




          1




          I added some lines is it more clear now ? @clathratus
          – Isham
          54 mins ago




          I added some lines is it more clear now ? @clathratus
          – Isham
          54 mins ago




          1




          1




          true @WillJagy I solved it for K positive I corrected my answer thanks a lot
          – Isham
          34 mins ago





          true @WillJagy I solved it for K positive I corrected my answer thanks a lot
          – Isham
          34 mins ago





          1




          1




          @WillJagy Thanks a lot for your answer I have upvoted it
          – Isham
          12 mins ago




          @WillJagy Thanks a lot for your answer I have upvoted it
          – Isham
          12 mins ago










          up vote
          1
          down vote













          as an example, in Isham's answer, if we choose $K$ so that
          $$ y' = frac12 (y^2-1), $$
          the first thing we notice is two stationary solutions, $y = 1$ and $y = -1$



          This is autonomous...



          For $y > 1 $ or $y < -1$ we get
          $$ y = frac-1tanh 2x $$
          which blow up in finite time, one in each direction. Also translates.



          For $-1 < y < 1$
          $$ y = - tanh 2x $$
          enter image description here






          share|cite|improve this answer


















          • 1




            @thanks a lot for your answer and the picture...that makes things more clear...
            – Isham
            4 mins ago














          up vote
          1
          down vote













          as an example, in Isham's answer, if we choose $K$ so that
          $$ y' = frac12 (y^2-1), $$
          the first thing we notice is two stationary solutions, $y = 1$ and $y = -1$



          This is autonomous...



          For $y > 1 $ or $y < -1$ we get
          $$ y = frac-1tanh 2x $$
          which blow up in finite time, one in each direction. Also translates.



          For $-1 < y < 1$
          $$ y = - tanh 2x $$
          enter image description here






          share|cite|improve this answer


















          • 1




            @thanks a lot for your answer and the picture...that makes things more clear...
            – Isham
            4 mins ago












          up vote
          1
          down vote










          up vote
          1
          down vote









          as an example, in Isham's answer, if we choose $K$ so that
          $$ y' = frac12 (y^2-1), $$
          the first thing we notice is two stationary solutions, $y = 1$ and $y = -1$



          This is autonomous...



          For $y > 1 $ or $y < -1$ we get
          $$ y = frac-1tanh 2x $$
          which blow up in finite time, one in each direction. Also translates.



          For $-1 < y < 1$
          $$ y = - tanh 2x $$
          enter image description here






          share|cite|improve this answer














          as an example, in Isham's answer, if we choose $K$ so that
          $$ y' = frac12 (y^2-1), $$
          the first thing we notice is two stationary solutions, $y = 1$ and $y = -1$



          This is autonomous...



          For $y > 1 $ or $y < -1$ we get
          $$ y = frac-1tanh 2x $$
          which blow up in finite time, one in each direction. Also translates.



          For $-1 < y < 1$
          $$ y = - tanh 2x $$
          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 15 mins ago

























          answered 33 mins ago









          Will Jagy

          98.3k595196




          98.3k595196







          • 1




            @thanks a lot for your answer and the picture...that makes things more clear...
            – Isham
            4 mins ago












          • 1




            @thanks a lot for your answer and the picture...that makes things more clear...
            – Isham
            4 mins ago







          1




          1




          @thanks a lot for your answer and the picture...that makes things more clear...
          – Isham
          4 mins ago




          @thanks a lot for your answer and the picture...that makes things more clear...
          – Isham
          4 mins ago

















           

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