A differential equation with a hidden sentence












10
















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



























    10
















    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

























      10












      10








      10









      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 Jan 1 at 12:36









      iBugiBug

      711219




      711219






















          2 Answers
          2






          active

          oldest

          votes


















          8














          This looks like:




          Happy New Year




          Update:



          Here's my solution:




          enter image description here







          share|improve this answer

































            4














            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
























            • G(rrr) = bear is also a nice stylistic choice of variables.

              – svavil
              Jan 2 at 8:30











            Your Answer





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8














            This looks like:




            Happy New Year




            Update:



            Here's my solution:




            enter image description here







            share|improve this answer






























              8














              This looks like:




              Happy New Year




              Update:



              Here's my solution:




              enter image description here







              share|improve this answer




























                8












                8








                8







                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 Jan 1 at 13:55

























                answered Jan 1 at 12:45









                piratepirate

                616118




                616118























                    4














                    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
























                    • G(rrr) = bear is also a nice stylistic choice of variables.

                      – svavil
                      Jan 2 at 8:30
















                    4














                    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
























                    • G(rrr) = bear is also a nice stylistic choice of variables.

                      – svavil
                      Jan 2 at 8:30














                    4












                    4








                    4







                    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 Jan 1 at 13:37









                    GlorfindelGlorfindel

                    13.6k34983




                    13.6k34983













                    • G(rrr) = bear is also a nice stylistic choice of variables.

                      – svavil
                      Jan 2 at 8:30



















                    • G(rrr) = bear is also a nice stylistic choice of variables.

                      – svavil
                      Jan 2 at 8:30

















                    G(rrr) = bear is also a nice stylistic choice of variables.

                    – svavil
                    Jan 2 at 8:30





                    G(rrr) = bear is also a nice stylistic choice of variables.

                    – svavil
                    Jan 2 at 8:30


















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