What does the substitution x=e^t do to \frac{dy}{dx}?

\frac{dy}{dx}=e^{-t}\frac{dy}{dt}, i.e. x\frac{dy}{dx}=\frac{dy}{dt}.

What is x^2\frac{d^2y}{dx^2} under x=e^t?

\frac{d^2y}{dt^2}-\frac{dy}{dt}.

Why does \frac{d^2y}{dx^2} contain a -\frac{dy}{dt} term?

The product rule: e^{-t} also depends on t, so differentiating e^{-t}\dot y gives e^{-t}(\ddot y-\dot y).

What type of equation does x=e^t solve?

An Euler equation ax^2y''+bxy'+cy=f(x).

How do you differentiate y=z\,e^{-x^2}?

Product rule: y'=(z'-2xz)e^{-x^2}.

After substituting y=zf(x), what common factor is cancelled?

The factor f(x) (such as e^{-x^2}), which is never zero.

How do you convert a condition on \frac{dy}{dx} to t?

Use \frac{dy}{dt}=x\frac{dy}{dx} (equal to \frac{dy}{dx} at x=1).

A particular integral has the same form as part of the complementary function. What do you do?

Multiply the trial form by t (by t^2 for a repeated root).

Second order equations reducible by substitutionEdexcel International A Level Further Maths: Flashcards

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What does the substitution $x=\mathrm{e}^t$ do to $\frac{dy}{dx}$?

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What does the substitution $x=\mathrm{e}^t$ do to $\frac{dy}{dx}$ ?: $\frac{dy}{dx}=\mathrm{e}^{-t}\frac{dy}{dt}$ , i.e. $x\frac{dy}{dx}=\frac{dy}{dt}$ .
What is $x^2\frac{d^2y}{dx^2}$ under $x=\mathrm{e}^t$ ?: $\frac{d^2y}{dt^2}-\frac{dy}{dt}$ .
What is $\frac{dt}{dx}$ when $x=\mathrm{e}^t$ ?: $\frac1x=\mathrm{e}^{-t}$ .
Why does $\frac{d^2y}{dx^2}$ contain a $-\frac{dy}{dt}$ term?: The product rule: $\mathrm{e}^{-t}$ also depends on $t$ , so differentiating $\mathrm{e}^{-t}\dot y$ gives $\mathrm{e}^{-t}(\ddot y-\dot y)$ .
What type of equation does $x=\mathrm{e}^t$ solve?: An Euler equation $ax^2y''+bxy'+cy=f(x)$ .
Under $x=\mathrm{e}^t$ , what does $ax^2y''+bxy'+cy=0$ become?: $a\ddot y+(b-a)\dot y+cy=0$ .
How do you differentiate $y=z\,\mathrm{e}^{-x^2}$ ?: Product rule: $y'=(z'-2xz)\mathrm{e}^{-x^2}$ .
After substituting $y=zf(x)$ , what common factor is cancelled?: The factor $f(x)$ (such as $\mathrm{e}^{-x^2}$ ), which is never zero.
What does a repeated auxiliary root $m$ give for the complementary function?: $(A+Bt)\mathrm{e}^{mt}$ .
Back-substitution: what is $\mathrm{e}^{-2t}$ in terms of $x$ ?: $\frac{1}{x^2}$ .
How do you convert a condition on $\frac{dy}{dx}$ to $t$ ?: Use $\frac{dy}{dt}=x\frac{dy}{dx}$ (equal to $\frac{dy}{dx}$ at $x=1$ ).
A particular integral has the same form as part of the complementary function. What do you do?: Multiply the trial form by $t$ (by $t^2$ for a repeated root).

Exam questions on Second order equations reducible by substitution

A differential equation for $y$ in terms of $x$ ( $x>0$ ) is to be transformed using the substitution $x=\mathrm{e}^t$ . A dot denotes differentiation with respect to $t$ , so $\dot y=\frac{dy}{dt}$ and $\ddot y=\frac{d^2y}{dt^2}$ .
The equation $x^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=0$ transforms into $\ddot y-3\dot y+2y=0$ . Find the general solution for $y$ in terms of $x$ .2 marks
The differential equation $\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+(4x^2+2)y=8\mathrm{e}^{-x^2}$ is to be transformed using the substitution $y=z\,\mathrm{e}^{-x^2}$ , where $z$ is a function of $x$ .
The equation transforms into $\frac{d^2z}{dx^2}=8$ . Find the general solution for $y$ in terms of $x$ .2 marks
The differential equation $x^2\frac{d^2y}{dx^2}+5x\frac{dy}{dx}+4y=0$ , $x>0$ , is to be solved using the substitution $x=\mathrm{e}^t$ .
Show that the substitution transforms the equation into $\frac{d^2y}{dt^2}+4\frac{dy}{dt}+4y=0$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).