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.


2 Answers 2


This looks like:

Happy New Year


Here's my solution:

enter image description here


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

  • $\begingroup$ G(rrr) = bear is also a nice stylistic choice of variables. $\endgroup$
    – svavil
    Jan 2, 2019 at 8:30

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