What does solving a differential equation involve?

Integration , working back from a derivative to the function itself. Which technique is needed depends on the equation, and for many of them the variables have to be separated before you can start.

True or False? A boundary condition has to describe the start of the situation.

False. It can describe any known pairing of the two variables, at any point at all. A particle coming to rest after a certain time gives a perfectly good boundary condition, and it is nowhere near the start.

True or False? Setting up a differential equation and solving it are the same skill.

False. Setting one up is a translation from words into notation; solving it is integration. A question can perfectly well ask you to formulate an equation without ever asking you to solve it.

Once you have the particular solution, what can you do with it?

Substitute any value of the independent variable to predict the quantity at that moment. The solution is an ordinary function, so a temperature after four minutes, or sales after three months, is only a substitution away.

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Differential Equations (Edexcel A Level Maths: Pure): Flashcards

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What is the difference between a first order and a second order differential equation?

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What is the difference between a first order and a second order differential equation?
A first order equation contains only first derivatives, such as $\frac{d y}{d x}$ .
A second order equation contains a second derivative, such as $\frac{d^{2} y}{d x^{2}}$ .
Define general solution.
The solution of a differential equation that still contains an arbitrary constant, written in the form $y = g (x) + c$ .
Since $c$ can take any value, it describes not one curve but a whole family of them.
Why does a differential equation have infinitely many solutions?
Because integrating introduces a constant that the equation itself does nothing to fix.
Every value of $c$ gives a curve with the right gradient function, so they all satisfy the equation and differ from one another only by a vertical shift.
What does solving a differential equation involve?
Integration, working back from a derivative to the function itself.
Which technique is needed depends on the equation, and for many of them the variables have to be separated before you can start.
True or False?
Every member of the family of solutions has the same gradient at a given value of $x$ .
True.
The curves differ only in the value of $c$ , which shifts them vertically without changing any gradient.
That is exactly why the differential equation cannot tell them apart: it describes the gradient and nothing else.
Define particular solution.
The single member of the family of solutions that satisfies a given extra piece of information.
Finding it means pinning down the one value of $c$ that fits, so the answer is one curve rather than infinitely many.
Define boundary condition.
A piece of extra information giving the value of one variable when the other is known, which is what fixes the constant of integration.
An example is $y = 4$ when $x = 0$ .
When is a boundary condition called an initial condition?
When it describes the situation at the start of a model or experiment.
That usually means $t = 0$ , so an initial condition is simply a boundary condition placed at the beginning of the timescale.
How many conditions are needed to find the particular solution of a differential equation?
One.
Integrating once introduces a single constant of integration, and one boundary condition is enough to fix its value.
Knowing $y = 4$ when $x = 0$ is exactly such a condition.
True or False?
A boundary condition has to describe the start of the situation.
False.
It can describe any known pairing of the two variables, at any point at all.
A particle coming to rest after a certain time gives a perfectly good boundary condition, and it is nowhere near the start.
What makes a differential equation separable?
It can be written as a product of a function of $x$ and a function of $y$ , that is $\frac{d y}{d x} = f (x) g (y)$ .
That form is what allows every $y$ term to be moved to one side and every $x$ term to the other.
Is $\frac{d y}{d x} = g (y)$ separable, when there is no $x$ in it at all?
Yes. Read it as $\frac{d y}{d x} = 1 \times g (y)$ , so that $f (x) = 1$ .
Separating then gives $\int \frac{1}{g (y)} d y = \int 1 d x$ , and the right-hand side simply integrates to $x$ .
What are the steps of separation of variables?
Get every $y$ term with the $d y$ on one side and every $x$ term with the $d x$ on the other, then integrate both sides.
Use any boundary condition to find the constant, and rearrange into whatever form the question asks for.
Complete the statement, filling in what is added after integrating both sides:
$\int \frac{1}{g (y)} d y = \int f (x) d x +______$
The completed statement is:
$\int \frac{1}{g (y)} d y = \int f (x) d x + c$
Only one constant is needed rather than one for each side, since two would immediately combine into a single overall constant anyway.
True or False?
Every first order differential equation can be solved by separating the variables.
False.
The right-hand side has to be a product of a function of $x$ and a function of $y$ before the two can come apart.
$\frac{d y}{d x} = x + y$ cannot be separated at all, and solving it needs a method beyond this course.
What phrase in a question tells you a differential equation is needed?
"Rate of change", which always signals a derivative term.
A rate measured against time gives $\frac{d V}{d t}$ or similar, so the phrase translates straight into notation.
How do you write "the rate of change of $y$ is directly proportional to $y$ " as an equation?
$\frac{d y}{d t} = k y$ , introducing a constant of proportionality $k$ .
Inversely proportional to $y$ would give $\frac{d y}{d t} = \frac{k}{y}$ instead.
What does the instruction formulate ask you to do?
Write the situation as an equation.
You may have to choose and define your own letters first, such as $V$ for a volume and $h$ for a height, since working that uses undefined letters cannot be followed.
A quantity is decreasing. What must its differential equation show?
A negative rate of change, so the equation reads $\frac{d y}{d t} = - k y$ with $k$ taken positive.
Cooling, decay and anything draining away all need that minus sign, and leaving it out models growth instead.
True or False?
Setting up a differential equation and solving it are the same skill.
False.
Setting one up is a translation from words into notation; solving it is integration.
A question can perfectly well ask you to formulate an equation without ever asking you to solve it.
After solving, a model gives $\ln y = k t + c$ . How do you put it in a useful form?
Take exponentials of both sides: $y = e^{k t + c} = e^{c} e^{k t}$ .
Since $e^{c}$ is just another constant, rename it $A$ , giving $y = A e^{k t}$ , the standard growth and decay form.
Once you have the particular solution, what can you do with it?
Substitute any value of the independent variable to predict the quantity at that moment.
The solution is an ordinary function, so a temperature after four minutes, or sales after three months, is only a substitution away.
A model gives $y = A e^{k t}$ . How do you find when $y$ reaches a given value?
Substitute that value and solve for $t$ , which needs logarithms because $t$ sits in the power.
It is the reverse of predicting the quantity at a known time, and the algebra runs the other way round.
A cooling model solves to $T = 20 + A e^{- k t}$ . What does that say about the long term?
That $T$ approaches $20$ without ever quite reaching it, since $A e^{- k t}$ tends to zero but is never zero.
The $20$ is the temperature of the surroundings, and reading it straight off the solution is what interpreting a model looks like.
True or False?
A differential equation model needs a boundary condition before it can be used to predict anything.
True.
Without one you have only the general solution, which is a whole family of curves rather than a single prediction.
Modelling questions therefore almost always supply one, most often the value at $t = 0$ .

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