What turns a general solution into a particular solution ?

A boundary condition , which is a known pair of values that the solution must satisfy, turns it into a particular solution. Substituting the condition produces an equation for the arbitrary constant, and geometrically that picks out the single curve of the family through the given point.

A LevelFurther MathsEdexcelCore PureFlashcardsDifferential EquationsFirst Order Differential Equations

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

Exam code: 9FM0

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FrontIntro to Differential Equations
Define differential equation.

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Cards in this collection (26)

Define differential equation.
A differential equation is any equation that contains at least one derivative term.
For example $\frac{d y}{d x} = 12 x y^{2}$ is one, and so is $\frac{d^{2} x}{d t^{2}} - 5 \frac{d x}{d t} + 7 x = 5 \sin t$ .
What fixes the order of a differential equation?
The order is the highest derivative that appears anywhere in the equation.
So $\frac{d^{2} x}{d t^{2}} - 5 \frac{d x}{d t} + 7 x = 5 \sin t$ is second order, and the first derivative sitting alongside makes no difference to that.
Complete the rule linking a differential equation's order to its general solution:
The general solution gains one arbitrary constant for each $______$ carried out, so a second order equation has $______$ constants and needs the same number of conditions to fix them.
The completed rule is:
The general solution gains one arbitrary constant for each integration carried out, so a second order equation has two constants and needs the same number of conditions to fix them.
What kind of object is the solution to a differential equation?
The solution is a function, not a number.
Solving $x^{2} - 4 = 0$ gives the two values $x = \pm 2$ , but solving $\frac{d y}{d x} = 4 x$ gives $y = 2 x^{2} + c$ , which is a whole curve.
Solving $\frac{d y}{d x} = 4 x$ gives $y = 2 x^{2} + c$ . Why is this called a family of solutions, and how are its members related?
The curves form a family because $c$ is unknown, so infinitely many of them satisfy the equation.
They are vertical translations of one another: $y = 2 x^{2}$ , $y = 2 x^{2} + 2$ and $y = 2 x^{2} + 4$ all have exactly the same shape.
True or False?
An initial condition is simply a boundary condition that describes the situation at the start.
True.
Both do the same job, which is to supply the information needed to fix the arbitrary constants in a general solution.
The word initial is used when the condition describes the start of a model or experiment, which is usually $t = 0$ .
What turns a general solution into a particular solution?
A boundary condition, which is a known pair of values that the solution must satisfy, turns it into a particular solution.
Substituting the condition produces an equation for the arbitrary constant, and geometrically that picks out the single curve of the family through the given point.
True or False?
A particle that is initially at rest has zero acceleration at $t = 0$ .
False.
Being at rest fixes the velocity only, giving the boundary condition $v = 0$ when $t = 0$ .
The acceleration at that instant can be anything: if $\frac{d v}{d t} = 2 t - 2$ then the particle starts at rest and yet its acceleration is $- 2 {m s}^{- 2}$ .
A particle is initially at rest and its velocity is later found to be $v = t^{2} - 2 t$ . Why is the second time it is at rest $t = 2$ and not $t = 0$ ?
Setting $v = 0$ gives $t (t - 2) = 0$ , so the particle is at rest at $t = 0$ and at $t = 2$ .
The particle was already at rest at $t = 0$ , so that root is the first occasion and $t = 2$ is the second.
Not every first order differential equation needs an integrating factor. Which ones can be integrated straight away?
Those already in the form $\frac{d y}{d x} = f (x)$ can be integrated directly, with no $y$ on the right-hand side to get in the way.
For example $\frac{d y}{d x} = 3 x^{2}$ gives $y = x^{3} + c$ in a single step.
True or False?
Every first order differential equation can be solved by separating the variables.
False.
Separation needs the equation to split into a function of $x$ multiplied by a function of $y$ , and many will not split that way.
$\frac{d y}{d x} + 6 y = e^{- 2 x}$ is one of them, which is exactly the gap an integrating factor fills.
Why must $\frac{d y}{d x} = 2 x y + 5 e^{x^{2}}$ be rearranged before its integrating factor can be found?
An integrating factor is only defined once the equation is in the standard form $\frac{d y}{d x} + p (x) y = q (x)$ , with the $y$ term on the left.
Moving $2 x y$ across gives $\frac{d y}{d x} - 2 x y = 5 e^{x^{2}}$ , so $p (x) = - 2 x$ and the minus sign is easily lost.
For an equation in the standard form $\frac{d y}{d x} + p (x) y = q (x)$ , fill in the index of the integrating factor:
$integrating factor = e^{______}$
The completed integrating factor is:
$integrating factor = e^{\int p (x) d x}$
Only $p (x)$ appears in it, so $q (x)$ plays no part in deciding which integrating factor to use.
Why is no constant of integration needed when finding an integrating factor?
Including one would multiply the integrating factor by a constant $e^{c}$ , and that constant then cancels from both sides of the equation.
Every choice of constant therefore gives a factor that works, so the simplest one is taken.
What are $p (x)$ and $q (x)$ for $\frac{d y}{d x} + 6 y = e^{- 2 x}$ ?
Here $p (x) = 6$ and $q (x) = e^{- 2 x}$ .
A function of $x$ is allowed to be a constant, which is easy to miss, and the integrating factor is then $e^{\int 6 d x} = e^{6 x}$ .
What does multiplying through by the integrating factor achieve?
Multiplying through turns the whole left-hand side into a single derivative of a product, so it can be integrated in one step.
The equation becomes $\frac{d}{d x} (y e^{\int p (x) d x}) = q (x) e^{\int p (x) d x}$ , and the left side integrates straight back to $y e^{\int p (x) d x}$ .
True or False?
Two integrations are carried out, so the solution ends up with two constants of integration.
False.
Only one constant appears, and it is introduced at the final integration of $q (x)$ multiplied by the integrating factor.
The earlier integration, the one sitting in the exponent, is taken without a constant at all.
After integrating both sides, a solution reads $y e^{- x^{2}} = 5 x + c$ . What is the final step, and where does $c$ end up?
Multiply through by $e^{x^{2}}$ to make $y$ the subject, giving $y = e^{x^{2}} (5 x + c)$ .
The constant stays inside the bracket, because it is multiplied by the exponential along with everything else rather than being added on at the end.
Why are differential equations a natural way to model real-world change?
A derivative is a rate of change, so an equation containing one is already a statement about how a quantity is changing.
Most real contexts are about change over time, so the derivative is usually taken with respect to $t$ .
What does the phrase rate of change of $P$ translate to in symbols?
The phrase becomes the derivative $\frac{d P}{d t}$ , taken with respect to time unless the context names another variable.
Wordings such as rate of growth of and rate of decay of mean the same thing.
The rate of change of a population of bacteria $P$ is proportional to the size of the population. Complete the model:
$\frac{d P}{d t} =______$
The completed model is:
$\frac{d P}{d t} = k P$
The constant of proportionality $k$ is not usually known at the outset, and is found later from the conditions supplied with the model.
The rate of change of the area $A$ covered by algae is proportional to the square root of that area. Write down the model.
The model is $\frac{d A}{d t} = k \sqrt{A}$ .
Proportionality need not be to the variable itself: whichever function of the variable the wording names is what goes on the right-hand side.
Why is a decreasing quantity usually modelled with $- k$ rather than by letting $k$ itself be negative?
Assuming $k > 0$ throughout and writing the minus sign explicitly keeps the direction of the change visible in the equation.
Both conventions describe the same model, but with $k > 0$ fixed you can tell growth from decay by looking at the sign rather than by working out what $k$ must be.
True or False?
A model can be written down and solved even though its constant of proportionality is unknown.
True.
The constant $k$ is simply carried through the working as an unknown, in much the same way as a constant of integration.
Conditions supplied with the model then give equations for $k$ and for the constant of integration together.
Newton's Law of Cooling makes the rate of change of an object's temperature $T$ proportional to $T - T_{a}$ , where $T_{a}$ is the ambient temperature. Why does the model carry a minus sign for an object that starts warmer than its surroundings?
The model is $\frac{d T}{d t} = - k (T - T_{a})$ with $k > 0$ .
Starting warmer makes $T - T_{a}$ positive, so the minus sign is what forces $\frac{d T}{d t}$ to be negative and the object to cool.
Water flows into a tank and also drains out of it. How does each flow appear in the model?
Anything increasing the quantity is added to the rate of change and anything decreasing it is subtracted.
So the model takes the shape $\frac{d V}{d t} = rate in - rate out$ , and any number of separate effects can be combined this way.

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