The proportionality of $d^2y/dx^2$ and $y$ 
Suppose that $x$ and $y$ are related by the equation $\displaystyle x=\int_0^y\frac{1}{\sqrt{1+4t^2}}\,dt$. Show that $d^2y/dx^2$ is proportional to $y$ and find the constant of proportionality.

I try to calculate the problem and I got the proportionality is $4(1+4y^2)^{-1/2}$
But it is not a constant.
Thank you so much.
 A: Here is a solution that gives explicitly the expression of functions we are working on.
You may know that 
$$\text{the antiderivative of} \ \dfrac{1}{\sqrt{1+t^2}} \ \text{is x=sinh}^{-1}(t)+k$$
where $k$ is an arbitrary constant (sinh$^{-1}$ being the inverse hyperbolic sine also denoted arcsinh). 
Thus, using an elementary change of variable, 
$$\text{the antiderivative of} \ \dfrac{1}{\sqrt{1+(2y)^2}} \ is \ x=\tfrac12 \text{sinh}^{-1}(2y)+k \tag{1}$$
As $x$ is expressed in terms of $y$, we are invited to reverse (1) for expressing $y$ as a function of $x$ ; doing this is easy : first, transform (1) into :
$$2(x-k)=\text{sinh}^{-1}(2y)$$
then take the $\sinh$ of LHS and RHS ; we get 
$$\sinh(2(x-k))=2y \ \iff \ $$

$$y=\tfrac12 \sinh(2(x-k))\tag{2}$$

From there, you can easily check that the second derivative of $y$ is $4$ times $y$.
Remark : Of course, the converse is not true. (2) is not, by far, the most general solution to the different and classical issue : 

Find all functions proportional to their second derivative, 

which has to be divided into two cases, according to the sign of this proportionality constant (of the form $k^2$ or $-k^2$) yielding two second order ordinary differential equations (ODE) :
$$\frac{d^2y}{dt^2}=k^2 y \ \ \text{and} \ \ \frac{d^2y}{dt^2}=-k^2 y \tag{3}$$ 
with resp. solutions :
$$y=A \text{sinh}(kx+B)  \ \ \text{and} \ \ y=A \sin(kx+B)$$
where $A$ and $B$ are arbitrary constants. 
(The second ODE in (3) is called the ODE of harmonic oscillator).
A: $$\frac{dy}{dx}=\left(\frac{dx}{dy}\right)^{-1}=\left(\frac{d}{dy}\int_{0}^y\frac{1}{\sqrt{1+4t^2}}\right)^{-1}=\sqrt{1+4y^2}$$
so
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\sqrt{1+4y^2}=\frac{4y\frac{dy}{dx}}{\sqrt{1+4y^2}}=4y$$
the proportionality is $4$.
A: The constant is $4$: $\frac {dx }{dy} =(1+4y^{2})^{-1/2}$ so $\frac {dy }{dx} =(1+4y^{2})^{1/2}$. This gives $\frac {d^{2}y }{dx^{2}} =4y (1+4y^{2})^{-1/2}\frac {dy }{dx}=4y$. 
