IB AA HL May 2026 P2 TZC Q12 [modified]

IB AA HL May 2026 P2 TZC Q12 [modified]

Junior College 2
15 marks
Adapted from IB May 2026 Examination for H2 Mathematics

The variables \(x\) and \(y\) satisfy the differential equation \(\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}\), where \(x,y>0\). It is given that \(y=2\) when \(x=1\).

  1. 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]
  2. Using the substitution \(u=xy\), show that the differential equation can be written as \(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{u^3}{x^2}\).[3]
  3. Hence, by separating the variables and using the given initial condition, solve the differential equation, giving \(y\) as a function of \(x\).[5]
  4. State the largest possible domain of this solution containing \(x=1\), and its corresponding range.[3]
  5. By considering the shape of the solution curve for \(1\leq x\leq1.05\) and the straight-line steps in Euler's method, state whether the approximation in part (a) is an overestimate or an underestimate. Give a reason.[1]

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Answer:(a) \(2.39\) (b) \(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{u^3}{x^2}\) (c) \(y=\frac2{\sqrt{x(8-7x)}}\) (d) domain \(0<x<\frac87\), range \(y\geq\frac{\sqrt7}{2}\) (e) underestimate

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