Euler estimate, the substitution u equals xy and the separable solution of a differential equation with its domain and range

For \(x>0\) and \(y>0\), the variables \(x\) and \(y\) are related by the differential equation \(\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}\), and \(y=2\) when \(x=1\).

Euler’s method gives a numerical estimate of a solution curve. Starting from a known point, it moves a short distance \(h\) in the \(x\)-direction along a straight line whose gradient is the value of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at that point.

Each step therefore increases the estimate of \(y\) by \(h\) multiplied by the gradient at the present point. Write \(y_n\) for the estimate of \(y\) when \(x=x_n\), where \(x_0=1\) and \(y_0=2\). Applied to this differential equation, the method is

\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]

The gradient is recalculated from the latest values of \(x\) and \(y\) before every new step, and unrounded values are carried forward throughout.

1. Let \(h=0.01\). Find the gradient at the point \((1,2)\), and use it in the formula above to show that the estimate at \(x_1=1.01\) is \(y_1=2.06\).
   Next find \(y_2\), showing your substitution, and give it correct to 5 decimal places. Repeat the process until you reach \(x=1.05\), and give your estimate of \(y\) there correct to 3 significant figures.[3]
2. By means of the substitution \(u=xy\), show that the equation can be rewritten as \(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{u^3}{x^2}\).[3]
3. Hence separate the variables and use the initial condition to solve the differential equation, expressing \(y\) in terms of \(x\).[5]
4. Write down the largest domain containing \(x=1\) on which this solution is valid, and the corresponding range.[3]
5. Considering how the solution curve bends on the interval \(1\leq x\leq1.05\), together with the straight-line steps that Euler’s method takes, state whether the estimate in part (a) overestimates or underestimates the true value. Justify your answer.[1]