HCA Further Mathematics: Differential Equations (2025 syllabus)

HCA Department of Mathematics Differential Equations Further Mathematics 9649 (2025 onward) Gerard Sayson Syllabus requirements ˆ Analytical solution of first order and second order linear differential equations of the form – dy dx + p(x)y = q(x) using an integrating factor – d2y dx2 + a dy dx + by = 0 for a, b∈ R – d2y dx2 + a dy dx + by = f (x) for a, b∈ R where f (x) is a polynomial or pekx or p cos(kx) or q sin(x) including those that can be reduced by means of a given substitution ˆ Relationship between the solution of a non-homogenous equation and the associated homogenous equation ˆ Family of solution curves ˆ Phase lines and slope fields (non-examinable) ˆ Exponential growth model ˆ Logistic growth model, equilibrium points and their stability, and har- vesting Document code: 9649-N-DE15 Class: P Published: 28 November 2024 Revision: N/A 1

Contents 1 Introduction 3 2 First order linear differential equations 5 2.1 Analytical solution for homogenous equations . . . . . . . . . 5 2.2 Analytical solution for non-homogenous equations . . . . . . 5 2.3 Alternate derivation of the solution . . . . . . . . . . . . . . . 7 3 Second order linear differential equations 9 3.1 Analytical solution for homogenous equations . . . . . . . . . 9 3.1.1 Distinct roots . . . . . . . . . . . . . . . . . . . . . . . 10 3.1.2 Repeated roots . . . . . . . . . . . . . . . . . . . . . . 11 3.2 Analytical solution for non-homogenous equations . . . . . . 12 3.2.1 Relationship with homogenous equations . . . . . . . 13 3.2.2 Method of undetermined coefficients . . . . . . . . . . 13 4 F amily of solution curves 15 4.1 Sketching solution curves . . . . . . . . . . . . . . . . . . . . 15 4.2 Phase lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.3 Slope fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 5 Population models 20 5.1 Exponential growth . . . . . . . . . . . . . . . . . . . . . . . . 20 5.2 Logistic growth . . . . . . . . . . . . . . . . . . . . . . . . . . 20 5.3 Harvesting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2

1 Introduction “Among all the mathematical disciplines, the theory of differential equations is the most important; it furnishes the explanation of all those elementary manifestations of nature which involve time.” Thus declared Sophus Lie, a famed mathematician responsible for ad- vances in linear and abstract algebra, and most importantly here, differential equations. Differential equations are responsible for virtually every single interac- tion in this universe (this is meant literally): it is the language of Physics and all the other physical sciences. Indeed, these kinds of equations allow us to model ongoing processes that evolve with time, like radioactive decay and population growth. So, what exactly is a differential equation? Differential equations are really just equations where the variable – usually a number in elementary algebra – is replaced by a derivative. It is quite similar to, but not exactly a polynomial. Here is an example: d2y dx2 + 3dy dx + 4y = 0 If we compare this to a polynomial, it would look something like: k2 + 3k + 4 = 0 As we can see, differential functions are made up of derivatives of varying orders. The main difference between regular, algebraic equations and differ- ential equations is that instead of solving for a value, we solve for a function. Hence we have the following: Definition 1.1. A linear ordinary differential equation of order n ∈ Z+ in y can be written in the form X 0≤k≤n fk(x) dky dxk = g(x) where y is a function ofx and d0y dx0 = y. Furthermore, the differential equation is called homogenous when g(x) = 0 and non-homogenous otherwise. Do note the mention of the wordordinary. This is for accuracy purposes; in fact, it is completely possible to form differential equations using partial derivatives. These equations are called partial differential equations , and they are outside the scope of the syllabus and this document. In H2 Mathematics, you will have dealt with differential equations of the form dy dx = f (x)g(y) 3

which can be solved by separating variables and integrating afterwards. If g(y) is not a linear polynomial (i.e. degree 1), then the equation is not linear. The Further Mathematics syllabus concerns itself with differential equations which are strictly linear, albeit harder to solve compared to the aforementioned method. 4

2 First order linear differential equations We recall that using Definition 1.1, a first order linear differential equation may be expressed in the form dy dx + P (x)y = Q(x) where P (x) and Q(x) are functions of x. Example 2.1. Consider the following equation: dy dx − 4x2y − 4 = 0 ⇐ ⇒dy dx − 4y = 4 This is a first order linear differential equation with P (x) = 4x2 and Q(x) = 4. This is a non-homogenous equation since Q(x) ̸= 0. 2.1 Analytical solution for homogenous equations We now proceed to find a solution to any equation in the form above. For starters, let us consider the general form of this type of equation: dy dx + P (x)y = Q(x) Clearly, we need some term in the left-hand side to contain Q(x). Hence y must contain Q(x) or some function of it. Now, let us integrate both sides to get rid of the derivative: Z dy dx + P (x)y

dx = Z Q(x) dx y + Z P (x)y dx= Z Q(x) dx We have y standing alone. Hence R P (x)y dxyields Q(x) in some way. Recall that integrating f ′(x)ef (x) with respect to x gives ef (x) + C. We could now think that maybe, P (x)y integrates to give a mere −y, when y = e− R P (x) dx. If only we could have Q(x) = 0, then indeed, y + Z P (x)y dx= y − Z (−P (x)) e− R P (x) dx dx = y − y = 0 2.2 Analytical solution for non-homogenous equations However, Q(x) isn’t always 0 for most equations. In that case, we need y to contain Q(x) as part of some factor, while containing e− R P (x) dx as a factor. 5

Much earlier before we began this chapter, we have learnt about the product rule. We can now start with y = e− R P (x) dxu(x) for some unknown function u(x), and get dy dx = e− R P (x) dx du dx − P (x)e− R P (x) dxu(x) We observe a common factor and thus factorize: dy dx = e− R P (x) dx du dx − P (x)u(x)

We substitute this inside our general equation: dy dx + P (x)y = Q(x) e− R P (x) dx du dx − P (x)u(x)

+ P (x)e− R P (x) dxu(x) = Q(x) e− R P (x) dx du dx = Q(x) Rearranging, we get du dx = Q(x)e R P (x) dx =⇒ u = Z Q(x)e R P (x) dx dx Hence, we have the following: Theorem 2.2. The solution of the equation dy dx + P (x)y = Q(x) is given by y = e− R P (x) dx Z Q(x)e R P (x) dx dx for any P (x) and Q(x). Proof. Consider the equation dy dx + P (x)y = Q(x) Differentiating y, one gets dy dx = −P (x)e− R P (x) dx Z Q(x)e R P (x) dx dx + e− R P (x) dxQ(x)e R P (x) dx = −P (x)e− R P (x) dx Z Q(x)e R P (x) dx dx + Q(x) 6

Also, P (x)y = P (x)e− R P (x) dx Z Q(x)e R P (x) dx dx Therefore we have dy dx + P (x)y = Q(x) and the proof is complete. Make sure to not forget your constant of integration! Example 2.3. Experiments indicate that the rate at which glucose is ab- sorbed by the body is λG, where λ ∈ R and G is the amount of glucose present in the bloodstream. Glucose is injected into a patient’s bloodstream at a constant rate of r units per unit time. Write a differential equation modelling the amount of glucose present in the patient’s bloodstream at time t. Hence, given that the initial amount of glucose present is G0, solve for G. Solution. We are given that the rate r is constant, and we have λG. We need to find a differential equation in G with respect to t. Furthermore, we are told that glucose is removed since it is absorbed into the body out of the bloodstream, so we must account for that. Hence we have dG dt = r − λG and so dG dt + λG = r G = e−λt r λ eλt + C

G = Ce−λt + r λ When t = 0, G = G0 so C =

G0 − r λ

. Hence G =

G0 − λ r

e−λt + r λ and we are done. 2.3 Alternate derivation of the solution @garbageskill remarks that there is a faster way with less guessing of what the solution