Differential Equations - Substitution

Solve the Initial Value Problem:
dy/dx = (3x + 2y)/(3x + 2y + 2) , y(-1) = -1

I need to use substitution and use the formulas:
du/dx = A + B(dy/dx)
dy/dx = (1/B)*(du/dx - A)

Let u = 3x + 2y then du/dx = 3 + 2(dy/dx)
therefore A = 3, B = 2

Pluging it into the formula gives me:
(1/2)*(du/dx - 3) = (3x + 2y)/(3x + 2y + 2)

Now I would need to separate the equation and integrate both sides, replace back what I substituted and solve the IVP where x = -1 and y = -1 to find the constant.

How do i make this equation a separable equation?

(1/2)*(du/dx - 3) = (u)/(u + 2)

Homework Statement

Solve the Initial Value Problem:
dy/dx = (3x + 2y)/(3x + 2y + 2) , y(-1) = -1

Homework Equations

I need to use substitution and use the formulas:
du/dx = A + B(dy/dx)
dy/dx = (1/B)*(du/dx - A)

The Attempt at a Solution

Let u = 3x + 2y then du/dx = 3 + 2(dy/dx)
therefore A = 3, B = 2

Pluging it into the formula gives me:
(1/2)*(du/dx - 3) = (3x + 2y)/(3x + 2y + 2)

Now I would need to separate the equation and integrate both sides, replace back what I substituted and solve the IVP where x = -1 and y = -1 to find the constant.

(1/2)*(du/dx - 3) = (u)/(u + 2)

(u+2)/u du = 3/2 *2 dx

integrate both sides to get:
u + 2 ln |u| = 3x + c
(3x + 2y) + 2 ln |(3x + 2y)| = 3x + c

plugging in the initial conditions I got:
-2 + 2 ln |5| = c

therefore,
(3x + 2y) + 2 ln |(3x + 2y)| = 3x - 2 + 2 ln |5|

is this correct?

tiny-tim