Home / A-Level Further Maths 9231 / Differential equations / Reducing equations with a substitution
Cambridge International AS & A Level · Topic 2.6
A-Level Further Maths 9231 / Topic 2.6 / AS & A Level

Change the variable, tame the equation.

Some differential equations look hopeless until you change a variable. Use \(y = ux\) to make an equation separable, \(x = e^{t}\) to turn an Euler equation into one with constant coefficients, and substitutions such as \(z = \frac{1}{y}\) or \(u = \frac{dy}{dx}\) to reach a first order linear equation. The substitution is always given: your job is the chain rule, the solving and the return to \(x\) and \(y\).

Topic 2.6All differential equations lessons →
The key idea

A given substitution swaps the hard equation for one you can already solve. With \(y = ux\): \(\frac{dy}{dx} = u + x\frac{du}{dx}\) (product rule), and the equation becomes separable in \(u\) and \(x\). With \(x = e^{t}\): \(x\frac{dy}{dx} = \frac{dy}{dt}\) and \(x^{2}\frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}\), so \(ax^{2}y'' + bxy' + cy\) becomes an equation with constant coefficients in \(t\). With \(z = y^{n}\) or \(u = \frac{dy}{dx}\): differentiate the substitution, replace every term, and look for \(\frac{dz}{dx} + P(x)z = Q(x)\). Every time, finish by writing the answer in the ORIGINAL variables, and only then use the initial conditions (or convert them carefully).

given substitution new equation solve with y = ux y′ = u + xu′ separable in u and x separate and integrate x = et xy′ = dy/dt x²y″ = d²y/dt² − dy/dt constant coefficients in t CF + PI, then t = ln x z = 1/y, z = y² or u = y′ first order linear integrating factor Last step every time: back to x and y, then use the conditions.
Fig. 1: the three families of given substitution in this lesson. Each turns a new equation into a type you have already met: separable, constant coefficients, or first order linear
Section 01

y = ux: making an equation separable.

An equation such as \(\frac{dy}{dx} = \frac{x^{2} + y^{2}}{xy}\) cannot be separated as it stands. Every term on the right has the same total degree in \(x\) and \(y\), so the right side depends only on the ratio \(\frac{y}{x}\). Putting \(y = ux\), where \(u\) is a function of \(x\), turns that ratio into a single new variable.

The one derivative you need comes from the product rule, because \(u\) and \(x\) both vary:

\[y = ux \;\Rightarrow\; \frac{dy}{dx} = u + x\frac{du}{dx}.\]

The method is always the same: replace \(y\) by \(ux\) and \(\frac{dy}{dx}\) by \(u + x\frac{du}{dx}\), cancel the powers of \(x\), take \(u\) to the right so that \(x\frac{du}{dx}\) stands alone, separate, integrate, and finally replace \(u\) by \(\frac{y}{x}\). Use a given condition on \(y\) at the very end, or turn it into a condition on \(u\) first (\(u = \frac{y}{x}\)).

Worked example A ratio on the right
Use the substitution \(y = ux\) to solve \(\frac{dy}{dx} = \frac{x^{2} + y^{2}}{xy}\) for \(x \gt 0\), given that \(y = 2\) when \(x = 1\).
  1. Replace \(y\) and \(\frac{dy}{dx}\): \[u + x\frac{du}{dx} = \frac{x^{2} + u^{2}x^{2}}{ux^{2}} = \frac{1 + u^{2}}{u}.\]
  2. Take \(u\) across: \[x\frac{du}{dx} = \frac{1 + u^{2}}{u} - u = \frac{1}{u}.\]
  3. Separate: \(\int u\,du = \int \frac{1}{x}\,dx\), so \(\tfrac{1}{2}u^{2} = \ln x + c\).
  4. Back-substitute \(u = \frac{y}{x}\): \(y^{2} = 2x^{2}\ln x + 2cx^{2}\).
  5. \(x = 1\), \(y = 2\): \(4 = 0 + 2c\), so \(2c = 4\).
Answer: \(y^{2} = x^{2}(2\ln x + 4)\)
Worked example An arctan appears
Use the substitution \(y = ux\) to solve \(x^{2}\frac{dy}{dx} = x^{2} + xy + y^{2}\) for \(x \gt 0\), given that \(y = 0\) when \(x = 1\). Give \(y\) in terms of \(x\).
  1. Divide by \(x^{2}\): \(\frac{dy}{dx} = 1 + \frac{y}{x} + \frac{y^{2}}{x^{2}}\), so \[u + x\frac{du}{dx} = 1 + u + u^{2}.\]
  2. The \(u\) terms cancel: \(x\frac{du}{dx} = 1 + u^{2}\).
  3. Separate: \(\int \frac{1}{1 + u^{2}}\,du = \int \frac{1}{x}\,dx\), so \(\tan^{-1}u = \ln x + c\).
  4. \(x = 1\), \(y = 0\) gives \(u = 0\), so \(0 = 0 + c\) and \(c = 0\).
  5. \(u = \tan(\ln x)\), and \(y = ux\).
Answer: \(y = x\tan(\ln x)\)
Your turn
Use \(y = ux\) to find the general solution of \(x\frac{dy}{dx} = x + 2y\) for \(x \gt 0\).
Examiner trap

Writing \(\frac{dy}{dx} = \frac{du}{dx}\), or \(\frac{dy}{dx} = x\frac{du}{dx}\), forgets the product rule: \(y = ux\) is a product of two functions of \(x\), so \(\frac{dy}{dx} = u + x\frac{du}{dx}\). And an answer left in \(u\) is unfinished: the question asked for \(y\).

Section 02

x = et: Euler equations.

An equation of the form \(ax^{2}\frac{d^{2}y}{dx^{2}} + bx\frac{dy}{dx} + cy = f(x)\) has variable coefficients, so the auxiliary equation method does not apply directly. The substitution \(x = e^{t}\) (for \(x \gt 0\)) makes every coefficient constant, because each power of \(x\) is matched by a derivative of the same order.

Derive the two results with the chain rule, using \(\frac{dx}{dt} = e^{t} = x\). The first derivative:

\[\frac{dy}{dt} = \frac{dy}{dx}\,\frac{dx}{dt} = x\frac{dy}{dx}.\]

Differentiate \(x\frac{dy}{dx}\) with respect to \(t\) by the product rule. The derivative of \(\frac{dy}{dx}\) with respect to \(t\) needs the chain rule again:

\[\begin{aligned} \frac{d^{2}y}{dt^{2}} &= \frac{d}{dt}\Big(x\frac{dy}{dx}\Big)\\ &= \frac{dx}{dt}\,\frac{dy}{dx} + x\,\frac{d}{dt}\Big(\frac{dy}{dx}\Big)\\ &= x\frac{dy}{dx} + x\,\frac{d^{2}y}{dx^{2}}\,\frac{dx}{dt}\\ &= x\frac{dy}{dx} + x^{2}\frac{d^{2}y}{dx^{2}}. \end{aligned}\]

Rearranging gives the pair to substitute:

\[x\frac{dy}{dx} = \frac{dy}{dt}, \qquad x^{2}\frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}.\]

Solve the new equation in \(t\) with a complementary function and a particular integral, then put \(t = \ln x\): \(e^{t} = x\), \(e^{2t} = x^{2}\), \(e^{-t} = \frac{1}{x}\), \(\cos(kt) = \cos(k\ln x)\). Initial conditions given at \(x = 1\) belong to \(t = 0\), and \(\frac{dy}{dt} = x\frac{dy}{dx}\) there, so the two derivatives happen to be equal at \(x = 1\) only.

Worked example Transform, then solve
Show that the substitution \(x = e^{t}\) transforms \(x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + 4y = 0\) into \(\frac{d^{2}y}{dt^{2}} + 4y = 0\). Hence find the general solution for \(y\) in terms of \(x\), for \(x \gt 0\).
  1. From the chain rule (derived above), \(x\frac{dy}{dx} = \frac{dy}{dt}\) and \(x^{2}\frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}\).
  2. Substitute: \[\Big(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}\Big) + \frac{dy}{dt} + 4y = 0,\] and the \(\frac{dy}{dt}\) terms cancel to give \(\frac{d^{2}y}{dt^{2}} + 4y = 0\).
  3. Auxiliary equation \(m^{2} + 4 = 0\), so \(m = \pm 2i\) and \(y = A\cos 2t + B\sin 2t\).
  4. Back-substitute \(t = \ln x\).
Answer: \(y = A\cos(2\ln x) + B\sin(2\ln x)\)
Worked example A particular solution with conditions
Use \(x = e^{t}\) to solve \(x^{2}\frac{d^{2}y}{dx^{2}} - 2x\frac{dy}{dx} + 2y = 6\ln x\) for \(x \gt 0\), given that \(y = \tfrac{5}{2}\) and \(\frac{dy}{dx} = 2\) when \(x = 1\).
  1. Substitute: \(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt} - 2\frac{dy}{dt} + 2y = 6t\), so \[\frac{d^{2}y}{dt^{2}} - 3\frac{dy}{dt} + 2y = 6t.\]
  2. CF: \(m^{2} - 3m + 2 = 0\), \(m = 1, 2\), so \(y_{c} = Ae^{t} + Be^{2t}\).
  3. PI: try \(y = pt + q\): \(-3p + 2pt + 2q = 6t\), so \(p = 3\) and \(2q = 9\), \(q = \tfrac{9}{2}\).
  4. Conditions: \(x = 1\) is \(t = 0\), and \(\frac{dy}{dt} = x\frac{dy}{dx} = 2\). Then \(A + B + \tfrac{9}{2} = \tfrac{5}{2}\) and \(A + 2B + 3 = 2\).
  5. So \(A + B = -2\) and \(A + 2B = -1\): \(B = 1\), \(A = -3\).
  6. \(y = -3e^{t} + e^{2t} + 3t + \tfrac{9}{2}\), and \(t = \ln x\) gives the answer.
Answer: \(y = x^{2} - 3x + 3\ln x + \tfrac{9}{2}\)
Your turn
Use \(x = e^{t}\) to find the general solution of \(x^{2}\frac{d^{2}y}{dx^{2}} + 3x\frac{dy}{dx} + y = x\) for \(x \gt 0\).
Examiner trap

Two slips cost the most marks. First, \(x^{2}\frac{d^{2}y}{dx^{2}}\) is NOT \(\frac{d^{2}y}{dt^{2}}\): the \(-\frac{dy}{dt}\) comes from the product rule and changes the auxiliary equation. Second, a right side such as \(6\ln x\) or \(x\) must also be rewritten in \(t\) (\(6t\), \(e^{t}\)) before you choose the particular integral; and the final answer goes back to \(x\).

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With the substitution \(y = ux\), where \(u\) is a function of \(x\), \(\frac{dy}{dx}\) equals

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With the substitution \(x = e^{t}\), the expression \(x\frac{dy}{dx}\) equals

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With the substitution \(x = e^{t}\), the expression \(x^{2}\frac{d^{2}y}{dx^{2}}\) equals

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The substitution \(x = e^{t}\) transforms \(x^{2}\frac{d^{2}y}{dx^{2}} + 4x\frac{dy}{dx} + 2y = 0\) into

Section 03

Other substitutions: reaching a linear equation.

Some given substitutions turn a non-linear equation into a first order linear one, \(\frac{dz}{dx} + P(x)z = Q(x)\), which an integrating factor then solves. Others change the dependent variable in a second order equation so that the coefficients become constant. The routine does not change: differentiate the substitution, replace every term, simplify, solve, return to \(y\).

Given substitutionWhat to replaceTypical result
\(z = \frac{1}{y}\)\(\frac{dz}{dx} = -\frac{1}{y^{2}}\frac{dy}{dx}\)first order linear in \(z\) (divide by \(y^{2}\) first)
\(z = y^{2}\)\(\frac{dz}{dx} = 2y\frac{dy}{dx}\)first order linear in \(z\)
\(u = \frac{dy}{dx}\)\(\frac{du}{dx} = \frac{d^{2}y}{dx^{2}}\)first order linear in \(u\), then integrate once more
\(y = \frac{z}{x}\)\(\frac{dy}{dx}\) and \(\frac{d^{2}y}{dx^{2}}\) by the quotient or product rulesecond order, constant coefficients in \(z\)

Each of these is the chain rule (or product rule) in disguise. Write the derivative of the substitution on its own line before you start replacing, so that no factor of \(y\) or \(x\) goes missing.

Worked example z = 1/y gives a linear equation
Use the substitution \(z = \frac{1}{y}\) to solve \(\frac{dy}{dx} + \frac{y}{x} = xy^{2}\) for \(x \gt 0\), given that \(y = 1\) when \(x = 1\).
  1. Divide by \(y^{2}\): \(\frac{1}{y^{2}}\frac{dy}{dx} + \frac{1}{x}\cdot\frac{1}{y} = x\).
  2. \(\frac{dz}{dx} = -\frac{1}{y^{2}}\frac{dy}{dx}\), so \(-\frac{dz}{dx} + \frac{z}{x} = x\), that is \[\frac{dz}{dx} - \frac{1}{x}z = -x.\]
  3. Integrating factor \(e^{-\ln x} = \frac{1}{x}\): \(\frac{d}{dx}\Big(\frac{z}{x}\Big) = -1\), so \(\frac{z}{x} = -x + c\) and \(z = cx - x^{2}\).
  4. \(x = 1\), \(y = 1\) means \(z = 1\): \(1 = c - 1\), so \(c = 2\).
  5. \(z = 2x - x^{2}\) and \(y = \frac{1}{z}\).
Answer: \(y = \dfrac{1}{2x - x^{2}}\)
Worked example y = z/x gives constant coefficients
Show that the substitution \(y = \frac{z}{x}\) transforms \(x\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + xy = 0\) into \(\frac{d^{2}z}{dx^{2}} + z = 0\), and hence find the general solution for \(y\).
  1. \(y = zx^{-1}\), so by the product rule \[\frac{dy}{dx} = \frac{1}{x}\frac{dz}{dx} - \frac{z}{x^{2}}.\]
  2. Differentiate again: \[\frac{d^{2}y}{dx^{2}} = \frac{1}{x}\frac{d^{2}z}{dx^{2}} - \frac{2}{x^{2}}\frac{dz}{dx} + \frac{2z}{x^{3}}.\]
  3. Then \(x\frac{d^{2}y}{dx^{2}} = \frac{d^{2}z}{dx^{2}} - \frac{2}{x}\frac{dz}{dx} + \frac{2z}{x^{2}}\) and \(2\frac{dy}{dx} = \frac{2}{x}\frac{dz}{dx} - \frac{2z}{x^{2}}\), while \(xy = z\).
  4. Add: every term with \(\frac{1}{x}\) cancels, leaving \(\frac{d^{2}z}{dx^{2}} + z = 0\).
  5. \(m^{2} + 1 = 0\) gives \(z = A\cos x + B\sin x\), and \(y = \frac{z}{x}\).
Answer: \(y = \dfrac{A\cos x + B\sin x}{x}\)
Your turn
Use \(u = \frac{dy}{dx}\) to find the general solution of \(x\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} = 12x^{2}\) for \(x \gt 0\).
Examiner trap

With \(z = \frac{1}{y}\), the minus sign in \(\frac{dz}{dx} = -\frac{1}{y^{2}}\frac{dy}{dx}\) is the mark-loser: drop it and the integrating factor is wrong from then on. With \(u = \frac{dy}{dx}\), solving for \(u\) is only halfway: you must integrate once more to reach \(y\), and the answer needs two constants.

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Using \(x = e^{t}\), the general solution of \(x^{2}\frac{d^{2}y}{dx^{2}} - 2x\frac{dy}{dx} + 2y = 0\), for \(x \gt 0\), is

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Use \(y = ux\) to find the general solution of \(\frac{dy}{dx} = \frac{x^{2} + 3y^{2}}{2xy}\), for \(x \gt 0\).

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The substitution \(z = \frac{1}{y}\) is applied to \(\frac{dy}{dx} + y = xy^{2}\). Which equation results?

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An Euler equation is solved with \(x = e^{t}\). The condition \(\frac{dy}{dx} = 3\) when \(x = 2\) becomes

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Six original Cambridge-style questions
A \(y = ux\) equation with an exponential, the derivation of the \(x = e^{t}\) results, a forced Euler equation with conditions, \(z = y^{2}\) and \(u = \frac{dy}{dx}\) reductions, and \(y = \frac{z}{x}\) with a limit condition.