Separating variables gives us:

`(dy)/y^3=(x\ dx)/(sqrt(1+4x^2))`

Integrating gives us:

`int(dy)/y^3=int(x\ dx)/(sqrt(1+4x^2)`

We now proceed to integrate the 2 sides separately. That is, we integrate the left side in *y* only (since after separating the variables we have terms in *y* and a *dy* on the left) and we work on the right side in *x* only (since we have terms in *x* and a *dx* only on the right).

For the right hand side involving *x*, let *u* = (1 + 4*x*^{2}), so *du* = 8*x* *dx* and *du*/8 = *x dx.*

`int(dy)/y^3=1/8int(du)/sqrtu`

`=(-1)/(2y^2)`

`=1/8(2)u^(1"/"2)+K`

`=1/4sqrt(1+4x^2)+K`

So the solution is given by:

`(-1)/(2y^2)=1/4sqrt(1+4x^2)+K`

We could go on to solve this in *y*, as follows:

Multiply both sides by `−2`:

`1/y^2=-1/2sqrt(1+4x^2)-2K`

For convenience, introduce a new variable `K_1 = -4K`, so that we'll have `-2K=K_1/2`. Our solution becomes:

`1/y^2=-1/2sqrt(1+4x^2)+K_1/2 =` ` (K_1 - sqrt(1+4x^2))/2`

Take the reciprocal of both sides:

`y^2=2/(K_1-sqrt(1+4x^2))`

Then solve for *y*:

`y=(+-sqrt2)/sqrt(K_1-sqrt(1+4x^2))`

(The constant *K*_{1} can be chosen so that the expression in the denominator is real.)