What is the point working it out if you already have the y value, is it just to get another y value at the end???

It's because when you integrate a function, the solution is a FAMILY of functions, due to there being an integration constant. The point of having a boundary condition is that you can solve for the integration constant.

In your case:

\(\displaystyle \frac{dy}{dx} = x\,y\,(1 - y)\)

\(\displaystyle \frac{1}{y\,(1 - y)}\,\frac{dy}{dx} = x\).

Using partial fractions:

\(\displaystyle \frac{A}{y} + \frac{B}{1 - y} = \frac{1}{y\,(1 - y)}\)

\(\displaystyle \frac{A(1 - y) + B\,y}{y\,(1 - y)} = \frac{1}{y\,(1 - y)}\)

\(\displaystyle A(1 - y) + B\,y = 1\)

\(\displaystyle (B - A)\,y + A = 0\,y + 1\).

Equating like coefficients gives

\(\displaystyle B - A = 0\) and \(\displaystyle A = 1\).

So \(\displaystyle B = 1\).

Therefore \(\displaystyle \frac{1}{y\,(1 - y)} = \frac{1}{y} + \frac{1}{1 - y}\).

Back to the DE:

\(\displaystyle \frac{1}{y\,(1 - y)}\,\frac{dy}{dx} = x\)

\(\displaystyle \left(\frac{1}{y} + \frac{1}{1 - y}\right)\,\frac{dy}{dx} = x\)

\(\displaystyle \int{\left(\frac{1}{y} + \frac{1}{1 - y}\right)\,\frac{dy}{dx}\,dx} = \int{x\,dx}\)

\(\displaystyle \int{\left(\frac{1}{y} + \frac{1}{1 - y}\right)\,dy} = \frac{1}{2}\,x^2 + C_1\)

\(\displaystyle \ln{|y|} - \ln{|1 - y|} + C_2 = \frac{1}{2}\,x^2 + C_1\)

\(\displaystyle \ln{\left|\frac{y}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\) where \(\displaystyle C = C_1 - C_2\)

\(\displaystyle \ln{\left|\frac{y - 1 + 1}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\)

\(\displaystyle \ln{\left|-\left(\frac{1 - y}{1 - y}\right) + \frac{1}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\)

\(\displaystyle \ln{\left|-1 + \frac{1}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\)

\(\displaystyle \left|-1 + \frac{1}{1 - y}\right| = e^{\frac{1}{2}\,x^2 + C}\)

\(\displaystyle \left|-1 + \frac{1}{1 - y}\right| = e^C\,e^{\frac{1}{2}\,x^2}\)

\(\displaystyle -1 + \frac{1}{1 - y} = A\,e^{\frac{1}{2}\,x^2}\) where \(\displaystyle A = \pm e^{C}\)

\(\displaystyle \frac{1}{1 - y} = A\,e^{\frac{1}{2}\,x^2} + 1\)

\(\displaystyle 1 - y = \frac{1}{A\,e^{\frac{1}{2}\,x^2}}\)

\(\displaystyle y = 1 - \frac{1}{A\,e^{\frac{1}{2}\,x^2}}\).

Now using the boundary condition that \(\displaystyle y(0) = \frac{1}{2}\) we find

\(\displaystyle \frac{1}{2} = 1 - \frac{1}{A\,e^{\frac{1}{2}(0)^2}}\)

\(\displaystyle \frac{1}{A\,e^{0}} = 1 - \frac{1}{2}\)

\(\displaystyle \frac{1}{A} = \frac{1}{2}\)

\(\displaystyle A = 2\).

So finally, the solution is

\(\displaystyle y = 1 - \frac{1}{2\,e^{\frac{1}{2}\,x^2}}\)

\(\displaystyle y = \frac{2\,e^{\frac{1}{2}\,x^2} - 1}{2\,e^{\frac{1}{2}\,x^2}}\).