Euler estimate, a reciprocal-square substitution and the particular solution of a first-order differential equation with its domain and range

A curve lies in the first quadrant, so that \(x>0\) and \(y>0\). It passes through the point where \(x=1\) and \(y=2\), and its gradient satisfies the differential equation \(\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}\).

1. Using Euler’s method with a step length of \(0.01\), estimate the value of \(y\) at \(x=1.05\).[3]
2. Let \(z\) be defined by \(\frac1{y^2}=-2z\).
   Show that \(z\) satisfies the differential equation \(\frac{\mathrm{d}z}{\mathrm{d}x}-\frac{2z}{x}=1\).[4]
3. Hence find the particular solution of the original differential equation, expressing \(y\) in terms of \(x\).[7]
4. State the domain of this particular solution and its range.[3]
5. With reference to the shape of the solution curve, say whether your estimate in part (a) overestimates or underestimates the true value. Give a reason for your answer.[1]