Substitution Methods for Solving Differential Equations

The key idea. The auxiliary-equation method only works for constant-coefficient equations $ay'' + by' + cy = f$. But some important equations have variable coefficients — most notably the Euler (equidimensional) equation $a x^2 \frac{d^2y}{dx^2} + b x \frac{dy}{dx} + c y = f(x),$ where each derivative is multiplied by a matching power of $x$. A substitution converts it into a constant-coefficient equation in a new variable, which you can then solve with the methods you already have.

The standard substitution is $x = e^t$, equivalently $t = \ln x$ (for $x > 0$). The magic is that the awkward $x^k \dfrac{d^k y}{dx^k}$ combinations become clean constant-coefficient derivatives in $t$.

The pattern, every time:

1. Transform the derivatives $\dfrac{dy}{dx}$, $\dfrac{d^2y}{dx^2}$ into $t$-derivatives via the chain rule.
2. Rewrite the ODE — it should collapse to constant coefficients in $t$.
3. Solve the standard equation (auxiliary equation for the CF, then a particular integral for $f$).
4. Back-substitute $t = \ln x$ to return to the original variable.

Cambridge tip. The question will tell you the substitution to use. Your job is to execute the derivative transforms accurately — that is where the marks are.

Everything depends on transforming $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ correctly. With $x = e^t$ we have $t = \ln x$ and $\dfrac{dt}{dx} = \dfrac{1}{x}$.

First derivative. By the chain rule, $\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{1}{x}\frac{dy}{dt} \quad\Rightarrow\quad x\frac{dy}{dx} = \frac{dy}{dt}.$

Second derivative. Differentiate $\dfrac{dy}{dx} = \dfrac{1}{x}\dfrac{dy}{dt}$ with respect to $x$ (product rule, and chain rule again on $\frac{dy}{dt}$): $\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x}\frac{d}{dx}\!\left(\frac{dy}{dt}\right) = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x}\cdot\frac{1}{x}\frac{d^2y}{dt^2}.$ Multiplying through by $x^2$: $\boxed{\,x^2\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}\,}$

These two boxed results are the only transforms you need to memorise for the $x = e^t$ substitution: $x\frac{dy}{dx} = \frac{dy}{dt}, \qquad x^2\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}.$

Notice how perfectly they fit the Euler equation: the $bx\,y'$ term becomes $b\,\dot y$ and the $ax^2y''$ term becomes $a(\ddot y - \dot y)$ — all constant coefficients.

Cambridge tip. A very common slip is writing $x^2\dfrac{d^2y}{dx^2} = \dfrac{d^2y}{dt^2}$ and dropping the $-\dfrac{dy}{dt}$. Derive (or quote) the full identity every time.

Once the equation is constant-coefficient in $t$, finish with the standard machinery and revert at the very end.

Step 1 — Complementary function. Solve the auxiliary equation of the homogeneous part in $t$. For example $\dfrac{d^2y}{dt^2} - \dfrac{dy}{dt} - 2y = 0$ has roots $m = 2, -1$, so $y_{\text{CF}} = Ae^{2t} + Be^{-t}$.

Step 2 — Particular integral. If $f(x)$ was, say, $x^3$, then $f = e^{3t}$ in the new variable — so try $y_{\text{PI}} = ke^{3t}$ and find $k$. (A right-hand side that was a power of $x$ becomes an exponential in $t$; a $\ln x$ becomes a polynomial in $t$.)

Step 3 — General solution in $t$, then back-substitute $t = \ln x$. Using $e^{kt} = e^{k\ln x} = x^k$ makes the reversion clean: $y = Ae^{2t} + Be^{-t} + ke^{3t} \;\xrightarrow{\,t = \ln x\,}\; y = Ax^2 + Bx^{-1} + kx^3.$ So the CF of an Euler equation is naturally a sum of powers of $x$ — which is exactly why $x = e^t$ is the right move.

Other substitutions, same discipline.

• $u = \dfrac{dy}{dx}$ (reduction of order): if $y$ is absent from the equation, set $u = y'$ so a second-order equation becomes first-order in $u$; solve for $u$, then integrate $u$ to recover $y$.
• $y = vx$ (homogeneous-type first-order): substitute $y = vx$, $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ to obtain a separable equation in $v$ and $x$.

In every case: transform, solve the standard equation, then undo the substitution.

Cambridge tip. Do not back-substitute too early. Carry the full $t$-solution (CF + PI) and replace $t = \ln x$ only once, at the end — reverting term-by-term mid-calculation invites errors.

Not every substitution is $x = e^t$. The syllabus expects you to apply any given substitution and transform the derivatives accordingly. The discipline is identical; only the algebra changes.

Reduction of order, $u = \dfrac{dy}{dx}$. Suppose $\dfrac{d^2y}{dx^2} + \dfrac{1}{x}\dfrac{dy}{dx} = 0$ (no $y$ term). Put $u = \dfrac{dy}{dx}$, so $\dfrac{du}{dx} = \dfrac{d^2y}{dx^2}$: $\frac{du}{dx} + \frac{1}{x}u = 0,$ a first-order separable/linear equation. Solve for $u$, then integrate once more to recover $y$ (adding a second constant). This is the standard trick when the dependent variable $y$ does not appear explicitly.

A substitution that linearises. If told to use $z = \dfrac{1}{y}$ or $z = y^2$ etc., differentiate the substitution to express $\dfrac{dy}{dx}$ in terms of $\dfrac{dz}{dx}$, substitute, and simplify to a standard form. For example with $z = y^{-1}$, $\dfrac{dz}{dx} = -y^{-2}\dfrac{dy}{dx}$.

The non-negotiable check. After substituting, verify that your transformed equation is genuinely standard (constant-coefficient, separable, or first-order linear) — if it is still messy, a derivative transform is almost certainly wrong. Re-derive the offending term.

Cambridge tip. Whatever the substitution, your final answer must be back in the original variables and must carry the correct number of arbitrary constants ($2$ for second-order, $1$ for first-order).