# A differential equation with a hidden sentence

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)

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.

This looks like:

Happy New Year

Update:

Here's my solution:

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

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