Differential equation involving function defined in terms of an integral 
Let $$F(y)=\int_0^\infty e^{-2x}\cos(2xy)\, dx$$ for $y\in \mathbb{R}$. Show that $F$ satisfies $F'(y)+2yF(y)=0$.

I've shown that I can differentiate under the integral sign, and this gives $F'(y)=-2\int_0^\infty xe^{-2x}\sin(2xy)\,dx$, but how in the world is this equal to $-2yF(y)$? Is there some trick I'm missing here?
 A: Integration by parts:
$$u=\sin 2xy\,\,,\,\,u'=2y\cos 2xy$$
$$v'=-xe^{-2x}\,\,,\,\,v=\frac{1}{2}e^{-2x}\,\,\,,\,\,\text{so:}$$
$$\left.-2\int_0^\infty xe^{-2x}\sin 2xy\,\,dx=\sin 2xy\,\,e^{-2x}\right|_0^\infty-2y\int_0^\infty e^{-2x}\cos 2xy\,\,dx$$
Now just check the first term in the RHS above is zero and you're done.
Added: The above is wrong and the reason why is in the comment by Adrian below, yet the solution is way more involved than it'd appear and I'll try to post it later.
Further added: I already tried but I can't get what the OP asks. Putting:
$$\mathcal I:=\int_0^\infty xe^{-2x}\sin 2xy\,\,dx\,\,\,,\,\,\,\mathcal J:=\int_0^\infty e^{-2x}\cos 2xy\,\,dx$$
I get the following integrating by parts (first time $\,u=x\sin 2xy\,$ , second time $\,\sin 2xy\,$ , third time $\,u=x\cos xy\,$ , and all the times $\,v'=e^{-2x}\,$ ): 
$$\mathcal I=\stackrel{\text{this equals zero}}{\overbrace{\left.-\frac{1}{2}xe^{-2x}\sin 2xy\right|_0^\infty}}+\frac{1}{2}\int_0^\infty e^{-2x}\sin 2xy\,\,dx+y\int_0^\infty xe^{-2x}\cos 2xy\,\,dx\Longrightarrow $$
$$\Longrightarrow \left.\mathcal I=-\frac{1}{4}e^{-2x}\sin 2xy\right|_0^\infty+\frac{y}{2}\mathcal J-\left.\frac{y}{2}xe^{-2x}\cos 2xy\right|_0^\infty+\frac{y}{2}\mathcal J-y^2\mathcal I$$
as the limits above equal zero, we get:
$$\mathcal I=yJ-y^2\mathcal I\Longrightarrow \mathcal I=\frac{y}{y^2+1}J$$
so I get that denominator $\,y^2+1\,$ there that shouldn't, according to the OP, be there. Either I'm mistaken or the OP is, but I already checked this several times. 
Everybody is wholeheartedly invited to either confirm my calculations of refute them. Thanks.
A: There is a mistake in the question posted. First note that, if we consider the integral

$$ \int_0^\infty e^{-sx}\cos(2xy)\, dx = {\frac {s}{{s}^{2}+4\,{y}^{2}}} $$

which gives the Laplace transform of $ \cos(2xy) $. Substituting $s=2$ gives

$$ F(y) = {\frac {2}{{4}+4\,{y}^{2}}} $$

Now, there is no way that $F(y)$ satisfies the given differential equation, trying this yields

$$  F'(y)+2yF(y)={\frac {{y}^{3}}{ \left( 1+{y}^{2} \right) ^{2}}} \neq 0 \,. $$

