Why do coupled equations arise so often in modelling?

Two quantities frequently affect each other's rate of change rather than only their own. In a predator and prey model the prey population depends both on how many prey there are to reproduce and on how many predators are eating them.

What is the overall strategy for solving a coupled pair?

Turn the pair into a single second order equation in one variable, which can then be solved by the standard method. Once that variable is known, the other is recovered by substitution rather than by solving a second differential equation.

True or False? A coupled model whose two solutions both grow exponentially for ever is unlikely to be realistic.

True. Unbounded growth ignores every practical limit, such as the food supply or the space available, so a model predicting it can be trusted only over a short interval. Real populations level off or oscillate instead, so a model of this kind is a reasonable short-term approximation and nothing more.

A LevelFurther MathsEdexcelCore PureFlashcardsDifferential EquationsSecond Order Differential Equations

Second Order Differential Equations (Edexcel A Level Further Maths: Core Pure): Flashcards

Exam code: 9FM0

1/19

0Still learning

Know0

FrontSolving Second Order Differential Equations
What makes a second order differential equation homogeneous?

Was this flashcard helpful?

Cards in this collection (19)

What makes a second order differential equation homogeneous?
The right-hand side is zero, so the equation reads $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = 0$ .
If the right-hand side is instead a non-zero function $f (x)$ the equation is non-homogeneous, and that difference decides how much work the solution takes.
Define coupled first order linear differential equations.
Coupled first order linear differential equations are a simultaneous pair, $\frac{d x}{d t} = a x + b y$ and $\frac{d y}{d t} = c x + d y$ , in which each derivative depends on both variables.
The word coupled names that mutual dependence: neither equation can be solved on its own.
True or False?
The general solution of $\frac{d^{2} y}{d x^{2}} = 0$ describes every possible non-vertical straight line.
True.
Integrating twice gives $y = A x + B$ , and choosing $A$ and $B$ produces any gradient and any intercept you please.
Vertical lines are the only ones missed, because $x = k$ has no $\frac{d y}{d x}$ at all.
Why do coupled equations arise so often in modelling?
Two quantities frequently affect each other's rate of change rather than only their own.
In a predator and prey model the prey population depends both on how many prey there are to reproduce and on how many predators are eating them.
Integrating $\frac{d^{2} y}{d x^{2}} = 24 x$ twice produces two constants. What becomes of the first one along the way?
The first constant becomes a term in $x$ , because the second integration acts on it as well.
Integrating once gives $\frac{d y}{d x} = 12 x^{2} + A$ , and integrating again gives $y = 4 x^{3} + A x + B$ , so only $B$ is left standing alone.
What is the overall strategy for solving a coupled pair?
Turn the pair into a single second order equation in one variable, which can then be solved by the standard method.
Once that variable is known, the other is recovered by substitution rather than by solving a second differential equation.
Complete the auxiliary equation belonging to $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = f (x)$ :
$______+ b m + c = 0$
The completed auxiliary equation is:
$a m^{2} + b m + c = 0$
It is an ordinary quadratic in $m$ , and its roots are what decide the shape of the complementary function.
Complete the route from a coupled pair to a single equation:
Rearrange one equation to make the other variable the subject, $______$ both sides with respect to $t$ , then substitute both results into the equation you did not use.
The completed route is:
Rearrange one equation to make the other variable the subject, differentiate both sides with respect to $t$ , then substitute both results into the equation you did not use.
Rearranging first is what makes the substitution possible, since the derivative of the second variable has to be available in terms of the first.
How do the roots of the auxiliary equation decide the complementary function?
There are three cases, according to the roots:
- two distinct real roots $α$ and $β$ give $y = A e^{α x} + B e^{β x}$
- a repeated root $α$ gives $y = (A + B x) e^{α x}$
- complex conjugate roots $p \pm q i$ give $y = e^{p x} (A \cos q x + B \sin q x)$
Having solved for $x$ and found $x = A e^{0.2 t} + B e^{0.3 t}$ , how do you find $y$ ?
Differentiate $x$ and substitute both $x$ and $\frac{d x}{d t}$ into the equation that was rearranged at the first step.
No new constants appear: the same $A$ and $B$ carry through into $y$ , and only the coefficients multiplying them change.
True or False?
The auxiliary equation depends on the function $f (x)$ on the right-hand side.
False.
The auxiliary equation is built only from the coefficients $a$ , $b$ and $c$ , so it is exactly the same whether $f (x)$ is zero or not.
That is why the complementary function can be found first and the right-hand side dealt with afterwards.
True or False?
A coupled model whose two solutions both grow exponentially for ever is unlikely to be realistic.
True.
Unbounded growth ignores every practical limit, such as the food supply or the space available, so a model predicting it can be trusted only over a short interval.
Real populations level off or oscillate instead, so a model of this kind is a reasonable short-term approximation and nothing more.
When the auxiliary equation has a repeated root $α$ , why is the complementary function $(A + B x) e^{α x}$ rather than $A e^{α x} + B e^{α x}$ ?
Those two terms collapse into one, since $A e^{α x} + B e^{α x} = (A + B) e^{α x}$ carries a single constant rather than two.
A second order equation needs two independent constants, and multiplying the second term by $x$ supplies one without simply repeating the first.
What do the signs of the exponents tell you about a coupled system's long-term behaviour?
A positive exponent makes its term grow without limit as $t$ increases, and a negative one makes its term decay away to zero.
Whichever exponent is largest eventually dominates, so that single term decides how the system behaves in the long run.
Complete the structure of the general solution of a non-homogeneous second order equation:
The general solution is the $______$ function added to the particular $______$ of the equation.
The completed structure is:
The general solution is the complementary function added to the particular integral of the equation.
For a homogeneous equation the particular integral is zero, so the complementary function is the whole general solution on its own.
What makes an expression a particular integral of a differential equation?
Substituting it into the left-hand side produces exactly $f (x)$ , the function sitting on the right.
Unlike a complementary function it carries no arbitrary constants, so on its own it is one specific solution rather than a family.
Complete two of the trial forms of particular integral:
For $f (x) = p e^{k x}$ try $λ______$ as the trial form, and for $f (x) = p \cos ω x + q \sin ω x$ try $λ \cos ω x + μ______$ instead.
The completed trial forms are:
For $f (x) = p e^{k x}$ try $λ e^{k x}$ as the trial form, and for $f (x) = p \cos ω x + q \sin ω x$ try $λ \cos ω x + μ \sin ω x$ instead.
The pattern holds throughout: the trial form has the shape of $f (x)$ with unknown constants in place of the given ones, so a polynomial gives a polynomial of the same degree and a constant gives a constant.
Why is the trial form for $\frac{d^{2} y}{d x^{2}} + 5 \frac{d y}{d x} + 6 y = 12 x$ taken as $λ + μ x$ rather than just $μ x$ ?
The full form is always used, even when a term is missing from $f (x)$ , because that term can still appear in the answer.
Here it does: comparing coefficients gives $μ = 2$ and $λ = - \frac{5}{3}$ , so the particular integral is $2 x - \frac{5}{3}$ even though the right-hand side has no constant term.
The equation $\frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + 3 y = 8 e^{3 x}$ has complementary function $A e^{x} + B e^{3 x}$ . Why will the trial form $λ e^{3 x}$ not work?
Every term of the complementary function makes the left-hand side zero, so $λ e^{3 x}$ can never produce $8 e^{3 x}$ .
Multiplying by $x$ escapes the clash: the trial form becomes $λ x e^{3 x}$ , and substituting it gives $λ = 4$ .

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:First Order Differential EquationsNext:Simple Harmonic Motion