Given the differential equation:

dy/dx = (y-1)/(x+3)

First we will group the terms of the same variable together.

We will multiply by dx.

==> dy = (y-1)/(x+3) * dx

Now we will divide by (y-1)

==> dy/(y-1) = dx/(x+3)

Now we will integrate both sides.

==> intg (dy/(y-1)) = intg dx/(x+3)

==> ln l y-1 l = ln l x+3l + C

Let us rewrite using the exponent form.

==> e^(ln l y-1l) = e^(ln lx+3l + C)

==> l y -1 l = e^ln l x+3l * e^C

==> l y-1 l = l x+ 3l * C1

==> l y-1 l = C1 * l x+3l

==> y-1 = +- C1 (x+3)

==> y = 1 + - C1 ( x+3)

Now we are given that y(-1) = 0

==> 0 = 1 +- C1 ( -1+3)

==> 0 = 1 +- C1 * 2

==> 0 = 1+-2C1

==> C1 = +-1/2

==> y = 1 +- (1/2) (x+3)

==> y= 1+- (x+3)/2

==> y = (2+(x+3)/2 = (x+5)/2

==> y= (2-x-3)/2 = -(x+1)/2

Then we have two possible values for y.

**y1= (x+5)/2**

**y2- -(x+1)/2**

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