Manipulate the ys on 1 side and xs on the other.

(2 + x)(dy/dx) = 3y
(dy/dx) = (3y) / (2 + x)
dy = (dx)(3y) / (2 + x)
(dy/3y) = (dx) / (2 + x)

Integrate both sides.
?(1/3)ydy = ?(1/(2 + x))dx
(1/6)y² + C = ln|2 + x| + C
(1/6)y² = ln|2 + x| + C

Note: a constant - a constant = still a constant.
(1/6)y² = ln|2 + x| + C

Plug in y(0) = 1 at this point, because you don't want to deal with square roots of an unknown constant, who know where you'll end up.

(1/6)(1)² = ln|2 + 0| + C
1/6 = ln(2) + C
C = (1/6) - ln(2)
C ˜ -0.526...

Plug C back into the your equation before you plugged in your initial conditon
(1/6)y² = ln|x + 2| + C
(1/6)y² = ln|x + 2| - 0.526
y² = 6ln|x + 2| - 3.156
y = v(6ln|x + 2| - 3.156)

Hope this helps :D

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