What is the difference between a general and a particular solution?

The general solution is the whole family of solutions, carrying an arbitrary constant like a constant of integration. A particular solution is one single member of that family, picked out by extra information.

True or False? A differential equation has exactly one solution.

False. There may be infinitely many solutions to a differential equation, one for each value of the arbitrary constant. Only a further condition, such as a point the solution passes through, picks out a single one.

Why are differential equations useful for modeling?

Because so many real quantities are described by their rate of change rather than directly, from a spreading infection to the drug level in a bloodstream. Writing that relationship as an equation and then solving it predicts how the quantity behaves.

What must a solution curve sketched on a slope field do?

It must pass through the given point and follow the general flow of the segments. It should not try to join up the grid points, because the solution is not known to pass through any of them.

True or False? A solution curve sketched on a slope field may run along a tangent segment.

True. Running along a segment is exactly what following the flow means, since the segment shows the solution's slope at that point. What the curve must never do is cut across a segment, which would give it a different slope from the one the equation dictates.

True or False? Integrating both sides of a separated equation needs two constants of integration.

False. One constant is enough, because the difference of two arbitrary constants is just another arbitrary constant. Write it on whichever side makes the rearranging easier.

True or False? More than one particular solution can pass through the same point.

False. The general solution is a family of curves, and exactly one member of it passes through any given point. That is why a single initial condition is enough to pin the solution down.

What does the definite integral form of a particular solution avoid?

Having to find a constant of integration at all. The lower limit and the starting value do that work instead.

Define Euler's method .

Euler's method approximates the solution to a first-order differential equation by taking short steps and treating the derivative as constant across each one. Each step is a linear approximation, using the tangent at the point you have reached.

How do you improve an Euler's method approximation?

Decrease the step size, which means using more steps to cover the same interval. An Euler approximation drifts away from the exact solution, and smaller steps keep it closer.

Why does a colony of bacteria fit an exponential growth model?

Because the more bacteria there are, the more new ones are produced, so the rate of change is proportional to the current size. A radioactive sample decays for the mirror reason: more atoms present means more atoms decaying.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 7: Differential EquationsFirst-Order Differential Equations

First-Order Differential Equations (College Board AP® Calculus BC): Flashcards

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FrontIntroduction to Differential Equations
Define a differential equation.

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Cards in this collection (56)

Define a differential equation.
A differential equation is an equation containing derivatives, so it relates a function to its own rates of change.
Both $\frac{d y}{d x} = 12 x y^{2}$ and $\frac{d^{2} x}{d t^{2}} - 5 \frac{d x}{d t} + 7 x = 5 \sin t$ are differential equations.
What makes a differential equation first order?
It contains first derivatives and no second or higher derivatives.
So $\frac{d y}{d x} = 12 x y^{2}$ is first order, while an equation containing $\frac{d^{2} x}{d t^{2}}$ is not.
What is the difference between a general and a particular solution?
The general solution is the whole family of solutions, carrying an arbitrary constant like a constant of integration.
A particular solution is one single member of that family, picked out by extra information.
True or False?
A differential equation has exactly one solution.
False.
There may be infinitely many solutions to a differential equation, one for each value of the arbitrary constant.
Only a further condition, such as a point the solution passes through, picks out a single one.
Fill in the two missing words about picking one solution out of the family:
a condition such as $y = 1$ when $x = 0$ is called an $______$ condition, and it determines the $______$ solution
The completed statement is: a condition such as $y = 1$ when $x = 0$ is called an initial condition, and it determines the particular solution.
It is also called a boundary condition, and graphically it names one point the solution curve must pass through.
Why are differential equations useful for modeling?
Because so many real quantities are described by their rate of change rather than directly, from a spreading infection to the drug level in a bloodstream.
Writing that relationship as an equation and then solving it predicts how the quantity behaves.
How do you verify that a proposed function solves a differential equation?
Differentiate it and check that the result matches the differential equation.
For $y = 1 - x - \cos x$ differentiating gives $- 1 + \sin x$ , which is $\sin x - 1$ , so it solves $\frac{d y}{d x} = \sin x - 1$ .
What extra technique does verifying an implicit solution need?
Implicit differentiation, since the proposed solution is not written as $y$ in terms of $x$ .
You may then also have to substitute the proposed solution back into your own result to reach the form the differential equation is written in.
Verify that $y^{2} = \frac{1}{x^{2} + C}$ solves $\frac{d y}{d x} = - x y^{3}$ .
Implicit differentiation of $y^{2} = {(x^{2} + C)}^{- 1}$ gives $2 y \frac{d y}{d x} = - \frac{2 x}{{(x^{2} + C)}^{2}}$ , so $\frac{d y}{d x} = - \frac{x}{y {(x^{2} + C)}^{2}}$ .
Since $\frac{1}{x^{2} + C} = y^{2}$ , that becomes $- \frac{x}{y} \cdot y^{4} = - x y^{3}$ as required.
Define a slope field.
A slope field is a diagram showing short tangent segments drawn at a grid of points in the plane.
The segment at any point has the slope that $\frac{d y}{d x}$ takes there, which is the slope of the solution curve through that point.
How do you find the slope to draw at the point $(x, y)$ ?
Write the differential equation as $\frac{d y}{d x} = g (x, y)$ , then substitute both coordinates into $g$ .
The number that comes out is the slope of the segment at that point.
Fill in the two missing words about what a slope field shows:
the segments act as $______$ lines, and a given point acts as a boundary condition naming which member of the $______$ of solutions to draw
The completed description is: the segments act as flow lines, and a given point acts as a boundary condition naming which member of the family of solutions to draw.
From any point the solution curve flows away in the direction the segment there is pointing.
What must a solution curve sketched on a slope field do?
It must pass through the given point and follow the general flow of the segments.
It should not try to join up the grid points, because the solution is not known to pass through any of them.
True or False?
A solution curve sketched on a slope field may run along a tangent segment.
True.
Running along a segment is exactly what following the flow means, since the segment shows the solution's slope at that point.
What the curve must never do is cut across a segment, which would give it a different slope from the one the equation dictates.
How do you find every point where the segments are horizontal?
Solve the equation $g (x, y) = 0$ , where $\frac{d y}{d x} = g (x, y)$ .
That finds them exactly, including any that fall between the grid points where segments were actually drawn.
For $\frac{d y}{d x} = - 0.4 {(y - 2)}^{\frac{1}{3}} (x - 1) e^{- \frac{{(x - 1)}^{2}}{25}}$ , where are the tangents horizontal?
Wherever $y = 2$ or $x = 1$ .
An exponential is never zero, so the product can only vanish when ${(y - 2)}^{\frac{1}{3}}$ or $(x - 1)$ does.
Does a horizontal segment mark a maximum or a minimum of a solution?
Not necessarily.
Every local maximum or minimum of a solution does sit where $\frac{d y}{d x} = 0$ , but not every such point is one.
Define separation of variables.
Separation of variables solves a first-order differential equation by getting all the $y$ terms on one side and all the $x$ terms on the other, then integrating both sides.
It works whenever the equation can be written as $\frac{d y}{d x} = g (x) h (y)$ .
Fill in the two missing differentials in the separated form of $\frac{d y}{d x} = g (x) h (y)$ :
$\frac{1}{h (y)}______= g (x)______$ ready for integrating both sides
The completed form is $\frac{1}{h (y)} d y = g (x) d x$ .
Every $y$ term ends up on one side and every $x$ term on the other, which is what makes each side integrable on its own.
Why is $\frac{d y}{d x} = 6 y$ suitable for separation of variables?
Because it is already a function of $x$ times a function of $y$ , with the $x$ part being the constant 6.
A constant counts as a function of $x$ , which is easy to miss.
True or False?
Integrating both sides of a separated equation needs two constants of integration.
False.
One constant is enough, because the difference of two arbitrary constants is just another arbitrary constant.
Write it on whichever side makes the rearranging easier.
Solve $\frac{d y}{d x} = \frac{e^{x} + 4 x}{3 y^{2}}$ .
It is $y = \sqrt[3]{e^{x} + 2 x^{2} + C}$ .
Separating gives $3 y^{2} d y = (e^{x} + 4 x) d x$ , which integrates to $y^{3} = e^{x} + 2 x^{2} + C$ .
Why can a general solution not always be written as $y = f (x)$ ?
Because taking a root or an inverse function throws away part of the solution.
From $y^{2} = \frac{1}{x^{2} + C}$ , writing only $y = \sqrt{\frac{1}{x^{2} + C}}$ loses the negative branch, so the squared form is left as it stands.
Does a fraction in a separated equation always integrate to a logarithm?
No.
A logarithm appears only when the integrand is the reciprocal of the variable itself, so $\int \frac{1}{y^{3}} d y = - \frac{1}{2 y^{2}} + C$ is an ordinary power.
What does an initial condition let you do?
Fix the value of the arbitrary constant in the general solution.
It is usually given as a value of $y$ for a particular value of $x$ , or as a point on the solution curve.
Fill in the two missing parts of the particular solution to $\frac{d y}{d x} = f (x)$ through the point $(a, y_{0})$ :
$F (x) =______+ \int_{a}^{x} f (______) d t$ for a dummy variable $t$
The completed solution is $F (x) = y_{0} + \int_{a}^{x} f (t) d t$ .
It satisfies the condition automatically, because $\int_{a}^{a} f (t) d t = 0$ leaves $F (a) = y_{0}$ .
True or False?
More than one particular solution can pass through the same point.
False.
The general solution is a family of curves, and exactly one member of it passes through any given point.
That is why a single initial condition is enough to pin the solution down.
What does the definite integral form of a particular solution avoid?
Having to find a constant of integration at all.
The lower limit and the starting value do that work instead.
Find the particular solution to $\frac{d y}{d x} = \frac{3}{x}$ given $y = 1$ when $x = e$ .
It is $y = 3 \ln | x | - 2$ .
Using $F (x) = 1 + \int_{e}^{x} \frac{3}{t} d t$ gives $1 + 3 (\ln | x | - 1)$ , since $\ln e = 1$ .
How do you find a particular solution once separation has given the general one?
Substitute the initial condition's values into the general solution and solve for the arbitrary constant.
There is no need to rearrange the general solution into $y = f (x)$ form first.
The general solution of $\frac{d y}{d x} = - x y^{3}$ is $y^{2} = \frac{1}{x^{2} + C}$ . Find $C$ if $y = \frac{1}{3}$ when $x = 2$ .
$C = 5$ .
Substituting gives $\frac{1}{9} = \frac{1}{4 + C}$ , so $4 + C = 9$ .
Define Euler's method.
Euler's method approximates the solution to a first-order differential equation by taking short steps and treating the derivative as constant across each one.
Each step is a linear approximation, using the tangent at the point you have reached.
What form must a differential equation be in for Euler's method?
It must be arranged as $\frac{d y}{d x} = f^{'} (x, y)$ , with the derivative alone on one side.
Sometimes only one of the two variables appears on the right, and the method runs in exactly the same way.
Fill in the two missing values in the Euler's method recursion:
$y_{n + 1} = y_{n} + Δ x \cdot f^{'} (______,______)$ where $Δ x$ is the step size
The completed recursion is $y_{n + 1} = y_{n} + Δ x \cdot f^{'} (x_{n}, y_{n})$ , alongside $x_{n + 1} = x_{n} + Δ x$ .
The derivative is evaluated at the point you are standing on, not at the one you are stepping to.
How do you work out the step size when it is not given?
Divide the total change in $x$ by the number of steps.
Going from $x = 2$ to $x = 3$ in two steps gives $Δ x = \frac{3 - 2}{2} = \frac{1}{2}$ .
Let $\frac{d y}{d x} = x - 3 y + 1$ with $f (0) = 2$ . Use two steps of $\frac{1}{2}$ to approximate $f (1)$ .
The approximation is $1$ .
The first step gives $y_{1} = 2 + \frac{1}{2} (0 - 6 + 1) = - \frac{1}{2}$ at $x = \frac{1}{2}$ .
The second gives $y_{2} = - \frac{1}{2} + \frac{1}{2} (\frac{1}{2} + \frac{3}{2} + 1) = 1$ .
True or False?
You must solve the differential equation before you can tell whether an Euler approximation is too big or too small.
False.
Differentiate $\frac{d y}{d x}$ again to get $\frac{d^{2} y}{d x^{2}}$ and look at its sign, with nothing to solve.
A concave down solution gives an overestimate and a concave up solution an underestimate.
$f (2) = 1$ , $f^{'} (2) = 5$ and $f^{'} (2.5) = 2$ . Use two equal steps to approximate $f (3)$ .
The approximation is $\frac{9}{2}$ .
With $Δ x = \frac{1}{2}$ the first step gives $1 + \frac{1}{2} (5) = \frac{7}{2}$ , and the second gives $\frac{7}{2} + \frac{1}{2} (2) = \frac{9}{2}$ .
How do you improve an Euler's method approximation?
Decrease the step size, which means using more steps to cover the same interval.
An Euler approximation drifts away from the exact solution, and smaller steps keep it closer.
Fill in the missing differential equation described by this statement:
the rate of change of a quantity is proportional to the size of the quantity, which is written as $______$ for a constant $k$
The completed model is $\frac{d y}{d t} = k y$ .
The constant $k$ is the constant of proportionality, and it is the phrase about the rate of change that tells you the model is exponential.
What does the sign of $k$ tell you in $\frac{d y}{d t} = k y$ ?
A positive $k$ means the quantity is growing and a negative $k$ means it is decaying.
Decay is sometimes written $\frac{d y}{d t} = - k y$ instead, with $k$ then taken to be positive.
Why does a colony of bacteria fit an exponential growth model?
Because the more bacteria there are, the more new ones are produced, so the rate of change is proportional to the current size.
A radioactive sample decays for the mirror reason: more atoms present means more atoms decaying.
Fill in the two missing parts of the solution to $\frac{d y}{d t} = k y$ with $y = y_{0}$ at $t = 0$ :
$y =______e^{______}$ for that same constant $k$
The completed solution is $y = y_{0} e^{k t}$ .
It is worth remembering outright, but it also follows from separating the variables and applying the initial condition.
How does separation of variables give $y = y_{0} e^{k t}$ ?
Separating $\frac{d y}{d t} = k y$ gives $\int \frac{1}{y} d y = \int k d t$ , so $\ln y = k t + C$ .
The initial condition makes $C = \ln y_{0}$ , and taking exponentials of both sides gives the result.
What are the doubling time and the half-life?
The doubling time is how long a growing quantity takes to double, and the half-life how long a decaying one takes to halve.
Both work out as $t = \frac{\ln 2}{k}$ .
True or False?
A larger starting amount gives a longer half-life.
False.
Solving $\frac{y_{0}}{2} = y_{0} e^{- k t}$ cancels $y_{0}$ entirely.
The half-life and the doubling time depend only on $k$ , never on how much there was to begin with.
A colony has $P = 5000$ at $t = 0$ and $P = 7000$ after 1 hour. When does it reach 100 000?
After about $8.903$ hours.
The model is $P = 5000 e^{k t}$ , and $7000 = 5000 e^{k}$ gives $k = \ln 1.4$ .
Then $e^{k t} = 20$ gives $t = \frac{\ln 20}{\ln 1.4}$ .
Fill in the missing differential equation described by this statement:
the rate of change of a quantity is jointly proportional to the size of the quantity and the difference between the quantity and the carrying capacity, which is written as $______$ for constants $k$ and $a$
The completed model is $\frac{d y}{d t} = k y (a - y)$ .
Here $a$ is the carrying capacity and $k$ sets how fast the quantity changes.
Define the carrying capacity in a logistic model.
The carrying capacity $a$ is the value the quantity converges towards as $t \to \infty$ .
It limits how large a growing quantity can become, and how small a shrinking one can get.
What does $\frac{d y}{d t} = k y (a - y)$ tell you when $y < a$ ?
That $\frac{d y}{d t} > 0$ , so the quantity is increasing towards $a$ .
When $y > a$ the derivative is negative and the quantity falls towards $a$ , and when $y = 0$ or $y = a$ it stays where it is.
True or False?
You must solve a logistic differential equation before you can find its carrying capacity.
False.
The carrying capacity can be read straight off $\frac{d y}{d t} = k y (a - y)$ as the non-zero value of $y$ that makes the derivative zero.
The value at which the quantity changes fastest can be found without solving it too.
At what value of $y$ is a logistic quantity changing fastest?
At $y = \frac{a}{2}$ , half the carrying capacity.
The expression $k y (a - y)$ is a downward quadratic in $y$ with zeros at $0$ and $a$ , so it peaks midway between them.
Fill in the two missing parts of the solution to the logistic equation when $0 < y_{0} < a$ :
$y = \frac{a}{1 + A e^{______}}$ where $A = \frac{a -______}{y_{0}}$ from the initial condition
The completed solution is $y = \frac{a}{1 + A e^{- a k t}}$ where $A = \frac{a - y_{0}}{y_{0}}$ .
The constant $A$ is simply the constant of integration rewritten.
Which two techniques are needed to solve a logistic differential equation?
Separation of variables first, and then partial fractions to integrate $\frac{1}{y (a - y)}$ .
That splits as $\frac{1}{a} (\frac{1}{y} + \frac{1}{a - y})$ , and each piece integrates to a logarithm.
A population satisfies $\frac{d N}{d t} = 0.0012 N (1500 - N)$ with $N = 300$ at $t = 0$ . Find $N$ in terms of $t$ .
It is $N = \frac{1500}{1 + 4 e^{- 1.8 t}}$ .
The carrying capacity is $1500$ and $a k = 1500 \times 0.0012 = 1.8$ , while $A = \frac{1500 - 300}{300} = 4$ .
For $N = \frac{1500}{1 + 4 e^{- 1.8 t}}$ , when is the population changing fastest?
At $t = \frac{\ln 4}{1.8}$ , about $0.770$ years.
The fastest change is at $N = 750$ , and solving $\frac{1500}{1 + 4 e^{- 1.8 t}} = 750$ gives $e^{- 1.8 t} = \frac{1}{4}$ .

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