A differential equation with a hidden sentence












2















Find the specialized solution of $H''(r)-aH'(r)=0$ that satisfies $H'(0) = frac{1}{p^2}$ and $H(0) = frac{H'(0)}{a}(1+frac{N}{Y}e^w)$




The question:




  1. Solve the above mathematic question

  2. Rewrite your solution to 1 in an appropriate form


  3. (Optional) This might be a hint (to some extents)




    Speak your answer to task 2 out loud!






Disclaimer: I am not the original author of this puzzle. It was an image that showed up in an instant messaging group. The original image contains solution for 1 and 2 and is written in my native language - Simplified Chinese.






mathematics steganography







share|improve this question



























    2















    Find the specialized solution of $H''(r)-aH'(r)=0$ that satisfies $H'(0) = frac{1}{p^2}$ and $H(0) = frac{H'(0)}{a}(1+frac{N}{Y}e^w)$




    The question:




    1. Solve the above mathematic question

    2. Rewrite your solution to 1 in an appropriate form


    3. (Optional) This might be a hint (to some extents)




      Speak your answer to task 2 out loud!






    Disclaimer: I am not the original author of this puzzle. It was an image that showed up in an instant messaging group. The original image contains solution for 1 and 2 and is written in my native language - Simplified Chinese.






    mathematics steganography







    share|improve this question

























      2












      2








      2








      Find the specialized solution of $H''(r)-aH'(r)=0$ that satisfies $H'(0) = frac{1}{p^2}$ and $H(0) = frac{H'(0)}{a}(1+frac{N}{Y}e^w)$




      The question:




      1. Solve the above mathematic question

      2. Rewrite your solution to 1 in an appropriate form


      3. (Optional) This might be a hint (to some extents)




        Speak your answer to task 2 out loud!






      Disclaimer: I am not the original author of this puzzle. It was an image that showed up in an instant messaging group. The original image contains solution for 1 and 2 and is written in my native language - Simplified Chinese.






      mathematics steganography







      share|improve this question














      Find the specialized solution of $H''(r)-aH'(r)=0$ that satisfies $H'(0) = frac{1}{p^2}$ and $H(0) = frac{H'(0)}{a}(1+frac{N}{Y}e^w)$




      The question:




      1. Solve the above mathematic question

      2. Rewrite your solution to 1 in an appropriate form


      3. (Optional) This might be a hint (to some extents)




        Speak your answer to task 2 out loud!






      Disclaimer: I am not the original author of this puzzle. It was an image that showed up in an instant messaging group. The original image contains solution for 1 and 2 and is written in my native language - Simplified Chinese.






      mathematics steganography




      mathematics steganography






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 hours ago









      iBug

      651118




      651118






















          2 Answers
          2






          active

          oldest

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          1














          Part 1:




          Start from $H''(r)-aH'(r)=0$ and let $G(r) = H'(r)$, so $G'(r)-aG(r)=0$. The general solution for this linear equation is $G(r) = be^{ar}$ ($b$ constant). Now $frac{1}{p^2} = H'(0) = G(0) = b$. Integrating $G(r)$ gives $H(r) = frac{e^{ar}}{a p^2}+C$ ($C$ constant). $H(0) = frac{1}{a p^2} + C$, so

          $$frac{1}{a p^2} + C = frac{H'(0)}{a}(1+frac{N}{Y}e^w) = frac{1}{a p^2}+frac{Ne^w}{a p^2 Y}$$


          so $C = frac{Ne^w}{a p^2 Y}$ and

          $$H = frac{e^{ar}}{a p^2}+frac{Ne^w}{a p^2 Y}$$




          Part 2:




          Rewriting this becomes

          $$H=frac{Ne^w}{a p p Y} + frac{Y e^{ar}}{a p p Y}$$


          and finally

          $$H a p p Y = N e^w + Y e^{a r}$$




          or, as @pirate correctly guessed,




          Happy New Year







          share|improve this answer





























            2














            This looks like:




            Happy New Year




            Update:



            Here's my solution:




            enter image description here







            share|improve this answer



















            • 1




              But where's your procedure? rot13(vg vf snveyl rnfl gb svther bhg guvf nafjre whfg ol ybbxvat ng gur yrggref)
              – iBug
              2 hours ago










            • It was a guess, I will work now on the proof ;)
              – pirate
              2 hours ago











            Your Answer





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            2 Answers
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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Part 1:




            Start from $H''(r)-aH'(r)=0$ and let $G(r) = H'(r)$, so $G'(r)-aG(r)=0$. The general solution for this linear equation is $G(r) = be^{ar}$ ($b$ constant). Now $frac{1}{p^2} = H'(0) = G(0) = b$. Integrating $G(r)$ gives $H(r) = frac{e^{ar}}{a p^2}+C$ ($C$ constant). $H(0) = frac{1}{a p^2} + C$, so

            $$frac{1}{a p^2} + C = frac{H'(0)}{a}(1+frac{N}{Y}e^w) = frac{1}{a p^2}+frac{Ne^w}{a p^2 Y}$$


            so $C = frac{Ne^w}{a p^2 Y}$ and

            $$H = frac{e^{ar}}{a p^2}+frac{Ne^w}{a p^2 Y}$$




            Part 2:




            Rewriting this becomes

            $$H=frac{Ne^w}{a p p Y} + frac{Y e^{ar}}{a p p Y}$$


            and finally

            $$H a p p Y = N e^w + Y e^{a r}$$




            or, as @pirate correctly guessed,




            Happy New Year







            share|improve this answer


























              1














              Part 1:




              Start from $H''(r)-aH'(r)=0$ and let $G(r) = H'(r)$, so $G'(r)-aG(r)=0$. The general solution for this linear equation is $G(r) = be^{ar}$ ($b$ constant). Now $frac{1}{p^2} = H'(0) = G(0) = b$. Integrating $G(r)$ gives $H(r) = frac{e^{ar}}{a p^2}+C$ ($C$ constant). $H(0) = frac{1}{a p^2} + C$, so

              $$frac{1}{a p^2} + C = frac{H'(0)}{a}(1+frac{N}{Y}e^w) = frac{1}{a p^2}+frac{Ne^w}{a p^2 Y}$$


              so $C = frac{Ne^w}{a p^2 Y}$ and

              $$H = frac{e^{ar}}{a p^2}+frac{Ne^w}{a p^2 Y}$$




              Part 2:




              Rewriting this becomes

              $$H=frac{Ne^w}{a p p Y} + frac{Y e^{ar}}{a p p Y}$$


              and finally

              $$H a p p Y = N e^w + Y e^{a r}$$




              or, as @pirate correctly guessed,




              Happy New Year







              share|improve this answer
























                1












                1








                1






                Part 1:




                Start from $H''(r)-aH'(r)=0$ and let $G(r) = H'(r)$, so $G'(r)-aG(r)=0$. The general solution for this linear equation is $G(r) = be^{ar}$ ($b$ constant). Now $frac{1}{p^2} = H'(0) = G(0) = b$. Integrating $G(r)$ gives $H(r) = frac{e^{ar}}{a p^2}+C$ ($C$ constant). $H(0) = frac{1}{a p^2} + C$, so

                $$frac{1}{a p^2} + C = frac{H'(0)}{a}(1+frac{N}{Y}e^w) = frac{1}{a p^2}+frac{Ne^w}{a p^2 Y}$$


                so $C = frac{Ne^w}{a p^2 Y}$ and

                $$H = frac{e^{ar}}{a p^2}+frac{Ne^w}{a p^2 Y}$$




                Part 2:




                Rewriting this becomes

                $$H=frac{Ne^w}{a p p Y} + frac{Y e^{ar}}{a p p Y}$$


                and finally

                $$H a p p Y = N e^w + Y e^{a r}$$




                or, as @pirate correctly guessed,




                Happy New Year







                share|improve this answer












                Part 1:




                Start from $H''(r)-aH'(r)=0$ and let $G(r) = H'(r)$, so $G'(r)-aG(r)=0$. The general solution for this linear equation is $G(r) = be^{ar}$ ($b$ constant). Now $frac{1}{p^2} = H'(0) = G(0) = b$. Integrating $G(r)$ gives $H(r) = frac{e^{ar}}{a p^2}+C$ ($C$ constant). $H(0) = frac{1}{a p^2} + C$, so

                $$frac{1}{a p^2} + C = frac{H'(0)}{a}(1+frac{N}{Y}e^w) = frac{1}{a p^2}+frac{Ne^w}{a p^2 Y}$$


                so $C = frac{Ne^w}{a p^2 Y}$ and

                $$H = frac{e^{ar}}{a p^2}+frac{Ne^w}{a p^2 Y}$$




                Part 2:




                Rewriting this becomes

                $$H=frac{Ne^w}{a p p Y} + frac{Y e^{ar}}{a p p Y}$$


                and finally

                $$H a p p Y = N e^w + Y e^{a r}$$




                or, as @pirate correctly guessed,




                Happy New Year








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 1 hour ago









                Glorfindel

                13.3k34982




                13.3k34982























                    2














                    This looks like:




                    Happy New Year




                    Update:



                    Here's my solution:




                    enter image description here







                    share|improve this answer



















                    • 1




                      But where's your procedure? rot13(vg vf snveyl rnfl gb svther bhg guvf nafjre whfg ol ybbxvat ng gur yrggref)
                      – iBug
                      2 hours ago










                    • It was a guess, I will work now on the proof ;)
                      – pirate
                      2 hours ago
















                    2














                    This looks like:




                    Happy New Year




                    Update:



                    Here's my solution:




                    enter image description here







                    share|improve this answer



















                    • 1




                      But where's your procedure? rot13(vg vf snveyl rnfl gb svther bhg guvf nafjre whfg ol ybbxvat ng gur yrggref)
                      – iBug
                      2 hours ago










                    • It was a guess, I will work now on the proof ;)
                      – pirate
                      2 hours ago














                    2












                    2








                    2






                    This looks like:




                    Happy New Year




                    Update:



                    Here's my solution:




                    enter image description here







                    share|improve this answer














                    This looks like:




                    Happy New Year




                    Update:



                    Here's my solution:




                    enter image description here








                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 54 mins ago

























                    answered 2 hours ago









                    pirate

                    2369




                    2369








                    • 1




                      But where's your procedure? rot13(vg vf snveyl rnfl gb svther bhg guvf nafjre whfg ol ybbxvat ng gur yrggref)
                      – iBug
                      2 hours ago










                    • It was a guess, I will work now on the proof ;)
                      – pirate
                      2 hours ago














                    • 1




                      But where's your procedure? rot13(vg vf snveyl rnfl gb svther bhg guvf nafjre whfg ol ybbxvat ng gur yrggref)
                      – iBug
                      2 hours ago










                    • It was a guess, I will work now on the proof ;)
                      – pirate
                      2 hours ago








                    1




                    1




                    But where's your procedure? rot13(vg vf snveyl rnfl gb svther bhg guvf nafjre whfg ol ybbxvat ng gur yrggref)
                    – iBug
                    2 hours ago




                    But where's your procedure? rot13(vg vf snveyl rnfl gb svther bhg guvf nafjre whfg ol ybbxvat ng gur yrggref)
                    – iBug
                    2 hours ago












                    It was a guess, I will work now on the proof ;)
                    – pirate
                    2 hours ago




                    It was a guess, I will work now on the proof ;)
                    – pirate
                    2 hours ago


















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