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AS & A Level · Practice questions · Reducing equations with a substitution

Substitute, solve, return.

Six original Cambridge-style questions on given substitutions in differential equations: \(y = ux\) to separable form, \(x = e^{t}\) for Euler equations, and substitutions that lead to first order linear or constant-coefficient equations, with particular solutions from initial conditions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Differentiate the substitution first.

01
Calculation
[5 marks]

Use the substitution \(y = ux\) to solve the differential equation \[x\frac{dy}{dx} = y + xe^{-y/x}, \quad x \gt 1,\] given that \(y = 0\) when \(x = 1\). Give \(y\) in terms of \(x\).

  • \(\frac{dy}{dx} = \frac{y}{x} + e^{-y/x}\), so \(u + x\frac{du}{dx} = u + e^{-u}\). ✓
  • The \(u\) terms cancel: \(x\frac{du}{dx} = e^{-u}\). ✓
  • Separate: \(\int e^{u}\,du = \int \frac{1}{x}\,dx\), so \(e^{u} = \ln x + c\). ✓
  • \(x = 1\), \(y = 0\) gives \(u = 0\): \(1 = 0 + c\), so \(c = 1\). ✓
  • \(u = \ln(1 + \ln x)\), so \(y = x\ln(1 + \ln x)\). ✓
02
Proof
[4 marks]

Given that \(x = e^{t}\), show that \[x\frac{dy}{dx} = \frac{dy}{dt} \quad\text{and}\quad x^{2}\frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}.\]

  • \(\frac{dx}{dt} = e^{t} = x\), so by the chain rule \(\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} = x\frac{dy}{dx}\). ✓
  • Differentiate \(\frac{dy}{dt} = x\frac{dy}{dx}\) with respect to \(t\), by the product rule: \[\frac{d^{2}y}{dt^{2}} = \frac{dx}{dt}\frac{dy}{dx} + x\frac{d}{dt}\Big(\frac{dy}{dx}\Big).\] ✓
  • By the chain rule, \(\frac{d}{dt}\big(\frac{dy}{dx}\big) = \frac{d^{2}y}{dx^{2}}\frac{dx}{dt} = x\frac{d^{2}y}{dx^{2}}\). ✓
  • So \(\frac{d^{2}y}{dt^{2}} = x\frac{dy}{dx} + x^{2}\frac{d^{2}y}{dx^{2}} = \frac{dy}{dt} + x^{2}\frac{d^{2}y}{dx^{2}}\), which rearranges to \(x^{2}\frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}\). ✓
03
Calculation
[7 marks]

Use the substitution \(x = e^{t}\) to solve \[x^{2}\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} - 3y = 4x, \quad x \gt 0,\] given that \(y = 1\) and \(\frac{dy}{dx} = 1\) when \(x = 1\).

  • Substitute: \(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt} - \frac{dy}{dt} - 3y = 4e^{t}\), so \[\frac{d^{2}y}{dt^{2}} - 2\frac{dy}{dt} - 3y = 4e^{t}.\] ✓
  • CF: \(m^{2} - 2m - 3 = 0\), \((m - 3)(m + 1) = 0\), so \(y_{c} = Ae^{3t} + Be^{-t}\). ✓
  • PI: try \(ke^{t}\): \(k - 2k - 3k = 4\), so \(-4k = 4\) and \(k = -1\). ✓
  • \(y = Ae^{3t} + Be^{-t} - e^{t}\). At \(x = 1\): \(t = 0\), \(y = 1\) and \(\frac{dy}{dt} = x\frac{dy}{dx} = 1\). ✓
  • \(A + B - 1 = 1\) and \(3A - B - 1 = 1\), so \(A + B = 2\) and \(3A - B = 2\): \(A = 1\), \(B = 1\). ✓
  • With \(t = \ln x\): \(y = x^{3} + \dfrac{1}{x} - x\). ✓
04
Calculation
[6 marks]

Use the substitution \(z = y^{2}\) to solve \[xy\frac{dy}{dx} + y^{2} = x^{2}, \quad x \gt 0,\] given that \(y = 1\) when \(x = 1\). Give \(y^{2}\) in terms of \(x\).

  • \(z = y^{2}\) gives \(\frac{dz}{dx} = 2y\frac{dy}{dx}\), so \(y\frac{dy}{dx} = \tfrac{1}{2}\frac{dz}{dx}\). ✓
  • The equation becomes \(\tfrac{1}{2}x\frac{dz}{dx} + z = x^{2}\), that is \[\frac{dz}{dx} + \frac{2}{x}z = 2x.\] ✓
  • Integrating factor \(e^{2\ln x} = x^{2}\): \(\frac{d}{dx}(x^{2}z) = 2x^{3}\), so \(x^{2}z = \tfrac{1}{2}x^{4} + c\). ✓
  • \(x = 1\), \(z = 1\): \(1 = \tfrac{1}{2} + c\), so \(c = \tfrac{1}{2}\). ✓
  • \(z = \dfrac{x^{4} + 1}{2x^{2}}\), so \(y^{2} = \dfrac{x^{4} + 1}{2x^{2}}\). ✓
05
Calculation
[6 marks]

Use the substitution \(u = \frac{dy}{dx}\) to solve \[x\frac{d^{2}y}{dx^{2}} - \frac{dy}{dx} = 3x^{2}, \quad x \gt 0,\] given that \(y = 2\) and \(\frac{dy}{dx} = 5\) when \(x = 1\).

  • \(\frac{du}{dx} = \frac{d^{2}y}{dx^{2}}\), so \(x\frac{du}{dx} - u = 3x^{2}\), that is \(\frac{du}{dx} - \frac{1}{x}u = 3x\). ✓
  • Integrating factor \(e^{-\ln x} = \frac{1}{x}\): \(\frac{d}{dx}\Big(\frac{u}{x}\Big) = 3\), so \(\frac{u}{x} = 3x + c\) and \(u = 3x^{2} + cx\). ✓
  • \(u = 5\) at \(x = 1\): \(3 + c = 5\), so \(c = 2\) and \(\frac{dy}{dx} = 3x^{2} + 2x\). ✓
  • Integrate: \(y = x^{3} + x^{2} + D\), and \(y = 2\) at \(x = 1\) gives \(D = 0\). ✓
  • \(y = x^{3} + x^{2}\). ✓
06
Analysis
[8 marks]

(a) Show that the substitution \(y = \frac{z}{x}\) transforms the differential equation \[x\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} - 4xy = 0, \quad x \gt 0,\] into \(\frac{d^{2}z}{dx^{2}} - 4z = 0\).

(b) Find the general solution for \(y\) in terms of \(x\).

(c) Given that \(y \to 0\) as \(x \to \infty\) and that \(y = 1\) when \(x = \tfrac{1}{2}\), find \(y\) in terms of \(x\).

  • (a) \(\frac{dy}{dx} = \frac{1}{x}\frac{dz}{dx} - \frac{z}{x^{2}}\) and \(\frac{d^{2}y}{dx^{2}} = \frac{1}{x}\frac{d^{2}z}{dx^{2}} - \frac{2}{x^{2}}\frac{dz}{dx} + \frac{2z}{x^{3}}\). ✓
  • So \[\begin{aligned} x\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} &= \frac{d^{2}z}{dx^{2}} - \frac{2}{x}\frac{dz}{dx} + \frac{2z}{x^{2}}\\ &\quad + \frac{2}{x}\frac{dz}{dx} - \frac{2z}{x^{2}}\\ &= \frac{d^{2}z}{dx^{2}}. \end{aligned}\] ✓
  • With \(4xy = 4z\), the equation becomes \(\frac{d^{2}z}{dx^{2}} - 4z = 0\), as required. ✓
  • (b) \(m^{2} - 4 = 0\), \(m = \pm 2\), so \(z = Ae^{2x} + Be^{-2x}\) and \(y = \dfrac{Ae^{2x} + Be^{-2x}}{x}\). ✓
  • (c) As \(x \to \infty\), \(\frac{e^{2x}}{x} \to \infty\), so \(y \to 0\) forces \(A = 0\); and \(\frac{Be^{-2x}}{x} \to 0\) for any \(B\). ✓
  • \(x = \tfrac{1}{2}\): \(1 = \dfrac{Be^{-1}}{1/2} = 2Be^{-1}\), so \(B = \tfrac{e}{2}\). ✓
  • \(y = \dfrac{e^{1 - 2x}}{2x}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.