This turns out to be a separable differential equation. We'll put all the y's on one side, and put all the x's on the other:

`dy/dx = 2y(x+2)/x`

Now, let's multiply both sides by `dx` and divide both sides by y:

`dy/y = 2(x+2)/x dx`

Now, we simply integrate both sides (we're going to use `C_1 and C_2` as our constants of integration) :

`int dy/y = 2 int (x+2)/x dx`

`lny + C_1 = 2(x + 2lnx + C_2)`

Now, let's simplify, recognizing that any manipulation of a constant of integration will be equivalent to another random constant, so we'l just let the resulting constant stay on the right side as "C":

`lny = 2x+4lnx +C`

Now, we let both sides be powers of e:

`e^(lny) = e^(2x + 4lnx + C)`

Recall that ln is the inverse function of the exponential function. Also, recall that adding exponents is equivalent to multiplication with another exponential function. These facts give us the following result when applied to the above equation:

`y = e^(2x)e^(4lnx) e^C`

Now, `4lnx = lnx^4` because of logarithm rules. Also, again because ln is the inverse function of e, we end up with the following result

`y = e^(2x) * x^4 * e^C`

Finally, we recognize that `e^C` is, you guess it, another random constant. So, let's call this one A (for fun) and simplify to get our final equation:

`y=Ax^4e^(2x)`

Because we are given no initial conditions, we cannot determine what A will be.

If you're not convinced that this function satisfies the above DE, we can substitute it in for y:

`dy/dx = 2y(x+2)/x`

`d/dx(Ax^4e^(2x)) = 2(Ax^4e^(2x))(x+2)/x`

To get the left-side term, we need to use the product and chain rules to get:

`A(4x^3e^(2x) + x^4(2e^(2x)))`

Simplifying:

`4Ax^3e^(2x) + 2Ax^4e^(2x)`

Now, let's simplify the right side of the equation:

`(2Ax^4e^(2x)*x + 2Ax^4e^(2x)*2)/x`

Simplifying:

`2Ax^4e^(2x) + 4Ax^3e^(2x)`

As you can see our final result is:

` ` `2Ax^4e^(2x) + 4Ax^3e^(2x) = 2Ax^4e^(2x) + 4Ax^3e^(2x)`

Therefore, our function solves the above D.E. See the link below for more info on ODE's.

Hope that helps!