2009 ACJC P1 Q9

The variables \(x\) and \(y\) are connected by the differential equation

\(\frac{\mathrm{d}y}{\mathrm{d}x}-1-{{\left( x+y \right)}^{2}}=0\).

1. Show that the substitution \(u=x+y\) reduces the differential equation to \(\frac{\mathrm{d}u}{\mathrm{d}x}-2-{{u}^{2}}=0\).[2]
2. Hence, solve \(\frac{\mathrm{d}y}{\mathrm{d}x}-1-{{\left( x+y \right)}^{2}}=0\), expressing \(y\) in terms of \(x\), given that \(y=2\) when \(x=0\).[4]
3. The variable \(A\) is defined to be the product of \(x\) and \(y\), where both \(x\) and \(y\)are functions of \(z\). If \(\frac{\mathrm{d}A}{\mathrm{d}z}=3\) when \(x=0\), find the value of \(\frac{\mathrm{d}x}{\mathrm{d}z}\) when \(x=0\).[3]