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 + 4x2), so du = 8x 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 K1 can be chosen so that the expression in the denominator is real.)