Why do we try y=e^{mx}?

Because y, y' and y'' are all multiples of e^{mx}, so the equation reduces to a quadratic in m.

Solution for complex roots \alpha\pm i\beta?

y=e^{\alpha x}(A\cos\beta x+B\sin\beta x)

What does a negative real part \alpha do?

The solution decays; with \beta\neq0, the oscillation has decaying amplitude.

How many conditions find A and B?

Two, e.g. y and \frac{dy}{dx} at x=0.

Second order equations with constant coefficientsAQA A-Level Further Maths: Flashcards

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Question

State the auxiliary equation for $y''+ay'+by=0$.

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State the auxiliary equation for $y''+ay'+by=0$ .: $m^2+am+b=0$
Why do we try $y=e^{mx}$ ?: Because $y$ , $y'$ and $y''$ are all multiples of $e^{mx}$ , so the equation reduces to a quadratic in $m$ .
Solution for two distinct real roots $m_1,m_2$ ?: $y=Ae^{m_1x}+Be^{m_2x}$
Solution for a repeated root $m$ ?: $y=(A+Bx)e^{mx}$
Solution for complex roots $\alpha\pm i\beta$ ?: $y=e^{\alpha x}(A\cos\beta x+B\sin\beta x)$
What discriminant gives distinct real roots?: $a^2-4b>0$
What discriminant gives a repeated root?: $a^2-4b=0$
What discriminant gives complex roots?: $a^2-4b<0$
General solution of $y''+9y=0$ ?: $y=A\cos3x+B\sin3x$
What does a negative real part $\alpha$ do?: The solution decays; with $\beta\neq0$ , the oscillation has decaying amplitude.
How many conditions find $A$ and $B$ ?: Two, e.g. $y$ and $\frac{dy}{dx}$ at $x=0$ .
Roots of $m^2+2m+5=0$ ?: $m=-1\pm2i$

Exam questions on Second order equations with constant coefficients

$\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0$ .
Find the particular solution for which $y=1$ and $\frac{dy}{dx}=0$ when $x=0$ .2 marks
$\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0$ .
Find the particular solution for which $y=2$ and $\frac{dy}{dx}=0$ when $x=0$ .2 marks
$\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=0$ .
Find the general solution.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).