A partial derivative question in a Chinese Calculus textbook Let $f(x)$ and $g(x)$ are derivable, and $u(x, y)=f(2x+5y)+g(2x-5y),\ u(x, 0)=\sin2x,\ u'_y(x,0)=0$. 
Find $u(x, y)=$?

This is a question in a Chinese Calculus textbook, and some possible answers should be: $\sin2x\cos5y,\ \sin(2x+5y),\ \sin2x+\sin5y,\ \sin2x-\sin5y$. But I don't know the correct answer. Thank you!
 A: If we know the choices you listed, then it should be: 

$(\sin2x\cos5y)'_{y=0}=-5\sin2x\sin5y=0,\ (\sin(2x+5y))'_{y=0}=5\cos(2x+5y)=5\cos2x,\ (\sin2x\pm\sin5y)'_{y=0}=\pm5\cos5y=\pm5$. 

So the correct answer is $u(x, y)=\sin2x\cos5y$.
A: So we know:
(1) $f(x)$ and $g(x)$ are derivable;
(2) $u(x,y)=f(2x+5y)+g(2x−5y)$;
(3) $u(x,0)=\sin2x$;
(4) $u′_y(x,0)=0$.
We should find $u(x,y)$.
The Solution: 
From (2) and (3) we'll have: $f(2x)+g(2x)=\sin2x$. (result A)
Since from (1) we know that $f(x)$ and $g(x)$ are derivable, then their sum $u = f + g$ shown at (2) will also be a derivable function. So let's calculate $u'_y(x, y)$:
$u'_y(x,y)=[f(2x+5y)+g(2x−5y)]'_y=[f(2x+5y)]'_y+[g(2x-5y)]'_y=$$=5f'_y(2x+5y)-5g'_y(2x−5y)$.  
Considering (4) on the latter we get:
$u'_y(x,0)=5f'_y(2x)-5g'_y(2x)=0$.
So we find that $f'_y(2x)-g'_y(2x)=0$, meaning $f(2x)-g(2x)=C$. (result B)
Here $C$ is a constant.  
From results A and B we get:  
$f(2x)=\frac{\sin2x+C}{2}$ and $g(2x)=\frac{sin2x-C}{2}$.  
Here by using (2) we can lastly conclude that:  
$u(x,y)=f(2x+5y)+g(2x−5y)=\frac{\sin(2x+5y)+C}{2}+\frac{\sin(2x-5y)-C}{2}=$
$=\frac{\sin(2x+5y)+\sin(2x-5y)}{2}=\frac{(\sin2x\cos5y+\cos2x\sin5y)+(\sin2x\cos5y-\cos2x\sin5y)}{2}=$
$=\sin2x\cos5y$.
