Euler's method estimates a solution of a differential equation by following the tangent direction in small steps. Let \((x_n,y_n)\) be the current estimated point on the solution curve, where \(y_n\) is an approximation to the value of \(y\) at \(x=x_n\).
The step size \(h\) is the increase in \(x\) between successive points: \(h=x_{n+1}-x_n\). It is a horizontal step, not the change in \(y\). For example, \(h=0.01\) gives the successive \(x\)-values \(1,\ 1.01,\ 1.02,\ldots\).
Since \(\frac{\mathrm{d}y}{\mathrm{d}x}\) gives the gradient, the change in \(y\) over a small horizontal step is approximated by \[\Delta y\approx h\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{(x_n,y_n)}.\] The vertical bar means that the derivative is evaluated by substituting the current values \(x=x_n\) and \(y=y_n\) into the differential equation.
Euler's method therefore defines the next estimated point by \[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{(x_n,y_n)}.\] This follows a straight line with the current gradient for one step. Recalculate the gradient at the new estimated point before taking the next step.
For the differential equation in this question, \[\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{(x_n,y_n)}=y_n^3-\frac{y_n}{x_n},\] so \(y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right)\).
Starting with \(x_0=1\) and \(y_0=2\), use \(h=0.01\) and five steps to estimate \(y\) when \(x=1.05\). Give your answer to 3 significant figures.[3]