2 Use the substitution \(z = x + y\) to find the solution of the differential equation $$\frac { d y } { d x } = \frac { 1 + 3 x + 3 y } { 3 x + 3 y - 1 }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\operatorname { aln } ( \mathrm { x } + \mathrm { y } ) + \mathrm { b } ( \mathrm { x } - \mathrm { y } ) + \mathrm { c } = 0\), where \(a , b\) and \(c\) are constants to be determined.