Solving various type of second order differential equation
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In many real life modelling situations, a differential equation for a variable of interest won’t just depend on the first derivative, but on higher ones as well. Naturally then, higher order differential equations arise in STEP and other advanced mathematics examinations. For anything more than a second derivative, the question will almost certainly be guiding you through some particular trick very specific to the problem at hand. For second order differential equations though, you need to know how to tackle them in general. Fortunately, the technique involved is straightforward, and this article guides you through all you need to know, with a helpful example as well!
Homogeneous Second Order Differential Equations
The first major type of second order differential equations you’ll have to learn to solve are ones that can be written for our dependent variable y and independent variable t as:
ad2ydt2+bdydt+cy=0.
Here a, b and c are just constants. In general the coefficients next to our derivatives may not be constant, but fortunately you don’t need to worry about how to approach such problems like that in general for STEP.
Now our approach to solving an equation of the above type is a simple one: we guess a solution. Of course, its an educated guess, there’s a lot of maths behind why we make the guess we do, but essentially it boils down to attempting a solution of the form y=eλt. Here, λ is simply an unknown constant, and our aim becomes to find λ for which a solution of this type satisfies the differential equation. Now, our guess implies that:
dydt=λeλt,d2ydt2=λ2eλt,
and we can therefore substitute into our differential equation to find:
aλ2eλt+bλeλt+ceλt=0.
Now because eλt is never zero, its safe to divide through to acquire a quadratic in λ:
aλ2+bλ+c=0.
This equation, which in future you can jump to straight away, is usually called the Auxiliary Equation.
But we know how to solve quadratics! This means we can find the λ for our a, b and c that allow eλt to satisfy our differential equation. Now, in general we’ll actually have 2 values for λ and our most general solution to the differential equation will be a linear combination of the two solutions they imply. Therefore, if we call our two solutions λ1 and λ2 we have:
y=Aeλ1t+Beλ2t.
But what happens if λ1=λ2? Well then instead we use:
y=(A+Bt)eλ1t.
Additionally, it’s important to realise that our λ may not necessarily be real numbers. If they happen to be complex, we could call our two solutions λ1=r+is and λ2=r−is, since they’ll always be complex conjugate pairs. Then our solution for y, using the relations between eit and the trigonometric functions, can be written as:
y=ert(Acosst+Bsinst).
So these three formula we’ve ended up with are all we actually need to remember! For any homogeneous second order differential equation with constant coefficients, we simply jump to the auxiliary equation, find our (\lambda\), write down the implied solution for y and then use initial conditions to help us find the constants if required.