Unit 7

Differential Equations: the one-page sheet

7.1 Writing Differential Equations

A differential equation states how fast a quantity changes. Translate words carefully: "the rate of change of y " is dy/dt , "proportional to" is " k times", and the sign of the right side tells you where y rises or falls. One known rate fixes k .

A differential equation relates a derivative to the variables: $\frac{dy}{dt} = 0.4(50 - y)$ .
"The rate of change of $y$ is proportional to $y$ " is $\frac{dy}{dt} = ky$ . The derivative is on the left; the quantity is on the right.
"Proportional to the difference between $T$ and 20" is $\frac{dT}{dt} = k(T - 20)$ .
The sign of $\frac{dy}{dt}$ says where $y$ increases or decreases; you can read it before knowing any formula for $y$ .
$k$ has units: for $\frac{dP}{dt} = kP$ with $t$ in hours, $k$ is "per hour".

write the rate as a derivative and the dependence as a formula: dP/dt = kP dy/dt = k⋅(that quantity) ; "inversely proportional" puts the quantity in the denominator the sign of dy/dt is known, so you know where the quantity grows and where it shrinks solve for k , then predict the rate at any other value

differential equation: An equation that contains the derivative of an unknown function, such as $\frac{dy}{dt} = 0.3y$ . It says how fast a quantity changes, often in terms of the quantity itself.
proportional to: " $A$ is proportional to $B$ " means $A = kB$ for a constant $k$ , the constant of proportionality. "Inversely proportional to $B$ " means $A = \frac{k}{B}$ .

7.2 Verifying Solutions to Differential Equations

To verify a solution to a differential equation, differentiate the function, substitute into both sides and show they agree for every x in an interval. One equation can have infinitely many solutions, and a solution may hold only on an interval where everything is defined.

A solution to a differential equation is a function, not a number. It makes the equation true for every $x$ in an interval.
To verify a solution: differentiate, substitute into both sides, simplify each side, compare.
Checking a single point is not enough: two different expressions can agree at one $x$ .
One equation can have a whole family of solutions, such as $y = Ce^{3x}$ for $\frac{dy}{dx} = 3y$ .
A solution may live on an interval only: $y = \frac{1}{x}$ solves $\frac{dy}{dx} = -y^2$ on $x > 0$ and on $x < 0$ , not across $x = 0$ .

compute every derivative the equation uses simplify each side separately, never moving terms across the = sign the function is a solution on that interval the function is not a solution there one differential equation usually has infinitely many solutions

solution to a differential equation: A function that makes a differential equation true for every $x$ in an interval: substitute it and its derivatives, and the two sides are equal. To verify a solution, differentiate, substitute and compare.

7.3 Sketching Slope Fields

A slope field draws dy/dx as a short segment at each point. Substitute the point's coordinates, get the slope, draw the segment. Look for where the slope is 0 or undefined, and for rows or columns of parallel segments.

At each point, compute $\frac{dy}{dx}$ and draw a short segment with that slope, centered on the point.
Slope 0 gives a horizontal segment; a slope that is undefined (division by 0) gives a vertical one.
If $\frac{dy}{dx}$ depends only on $x$ , segments in each vertical column are parallel. If it depends only on $y$ , segments in each horizontal row are parallel.
Find where the slope is 0 first: those lines organize the whole field.
Make steep slopes look steep and gentle ones gentle; the sign matters most.

each point of the plane gets a slope you get the number the segment through that point must have as its slope positive slopes rise to the right, negative ones fall, zero is flat the field shows how every solution curve bends

slope field: A picture of a differential equation $\frac{dy}{dx} = F(x, y)$ : at each chosen point $(x, y)$ , a short segment with slope $F(x, y)$ . A solution curve through a point follows the segments near it.

7.4 Reasoning with Slope Fields

Trace a solution through a point by following the segments across the field. Rows of flat segments are equilibrium solutions, which other solutions approach or leave but never cross. The sign of dy/dx gives direction; d²y/dx² , found with the chain rule, gives concavity.

Sketch a solution through a point by following the segments both ways, all the way across the field; never cut across the segments.
An equilibrium solution $y = c$ makes $\frac{dy}{dx} = 0$ for every $x$ . Solve "right side $= 0$ " for $y$ .
The sign of $\frac{dy}{dx}$ says where a solution rises or falls; long-run behavior shows whether curves approach an equilibrium or move away.
Concavity: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ . Every $y$ is a function of $x$ , so $\frac{d}{dx}(y) = \frac{dy}{dx}$ ; then substitute the equation.

it runs along the segment there and keeps following the segments in both directions y = c is an equilibrium solution, a row of flat segments a curve that starts below an equilibrium line stays below it its sign at a point tells the concavity of the solution there

equilibrium solution: A constant function $y = c$ that solves a differential equation: $\frac{dy}{dx} = 0$ at every point of the horizontal line $y = c$ . In a slope field it is a row of horizontal segments.

7.5 Euler's Method

Euler's method approximates a solution of dy/dx = F(x, y) by short tangent steps: from a known point, y_new = y + F(x, y) Δx , then repeat from the new point. Smaller steps follow the curve more closely. The sign of d²y/dx² tells whether the steps fall below the solution (concave up, an underestimate) or above it (concave down, an overestimate).

One step: $y_{\text{new}}$ $= y + \frac{dy}{dx}\Big|_{(x, y)} \cdot \Delta x$ , using the slope at the point you are at.
Lay the work out in a table with columns $x$ , $y$ , $\frac{dy}{dx}$ , $\Delta x$ , $\Delta y$ ; take exactly the steps asked for.
Smaller steps give better estimates. A negative $\Delta x$ steps to the left.
Over or under: find $\frac{d^2y}{dx^2}$ by differentiating the equation. Concave up curves lie above their tangent steps (underestimate); concave down, below (overestimate).

at a known point you know the direction of the solution curve the rise is Δy = dy/dx Δx , and the new point is (x + Δx, y + Δy) take the next step; repeat until you reach the x you want each step falls below (or above) them, so the estimate is too small (or too large)

Euler's method: A step-by-step approximation of a solution of $\frac{dy}{dx} = F(x, y)$ : from a known point $(x, y)$ , move to $(x + \Delta x,\ y + F(x, y)\,\Delta x)$ , then repeat from the new point with its own slope. Each step follows a tangent line; smaller steps give closer estimates.

7.6 General Solutions by Separation of Variables

For a separable equation, put y with dy and x with dx , integrate both sides, and add one constant immediately. Keep ln |y| ; when you exponentiate, the constant absorbs the sign. The general solution is a family of curves, one for each constant.

Separable means the right side is a product (or quotient) of a function of $x$ and a function of $y$ . $\frac{dy}{dx} = x + y$ is not separable.
Separate completely: only $y$ and $dy$ on one side, only $x$ and $dx$ on the other. $dy$ and $dx$ stay in the numerators.
Integrate both sides and write $+ C$ once, right away. Adding it after solving for $y$ gives a different, wrong family.
$\int \frac{dy}{y} = \ln|y|$ . Then $|y| = e^{C}e^{\dots}$ , and $\pm e^{C}$ becomes a constant $A$ that may be any sign.

the equation is separable: 1/(h(y)) dy = g(x) dx each side becomes an antiderivative; the two constants combine into one C keep the absolute value; exponentiating gives |y| = e^Ce^(G(x)) , so y = Ae^(G(x)) with A any constant the general solution: a family of curves, one for each constant

separation of variables: A way to solve $\frac{dy}{dx} = g(x)h(y)$ : move every $y$ to the side with $dy$ and every $x$ to the side with $dx$ , $\frac{1}{h(y)}\,dy = g(x)\,dx$ , then integrate both sides and add one constant $C$ . The result, a family with $C$ in it, is the general solution.

7.7 Particular Solutions from Initial Conditions

A particular solution satisfies the differential equation and an initial condition. Separate, integrate, add C , substitute the initial point immediately, then solve for y with the sign that fits. Give its domain: the open interval around the initial x where it is defined and differentiable.

An initial condition $y(x_0) = y_0$ says the solution passes through $(x_0, y_0)$ . Put $x_0$ in for $x$ and $y_0$ in for $y$ .
Use it as soon as $C$ appears. Solving for $y$ first invites algebra slips and wrong constants.
After a square root, keep the sign that matches $y_0$ : $y(1) = -3$ needs the negative root.
State the domain: an open interval containing $x_0$ on which the formula is defined and its derivative exists.

an equation with one constant C a number for C , before any rearranging one function, the particular solution the domain is the largest open interval around the initial x where the solution is defined and differentiable

particular solution: The one member of a general solution that satisfies an initial condition such as $y(0) = 1$ (a point the curve must pass through), together with its domain: the largest open interval containing the initial $x$ on which the solution is defined and differentiable.

7.8 Exponential Models

When the rate of change is proportional to the amount, dy/dt = ky and y = y_0e^kt . k > 0 is growth, k < 0 decay; k is the rate as a fraction of the amount per unit time. Find k by dividing two values and taking ln ; doubling time and half-life are (ln 2)/(|k|) .

$\frac{dy}{dt} = ky$ and $y(0) = y_0$ give $y = y_0e^{kt}$ . You may quote this result.
$k = \frac{dy/dt}{y}$ : the rate of change as a fraction of the current amount, in "per unit time" (per hour, per year).
Doubling time ( $k > 0$ ) and half-life ( $k < 0$ ) are both $\frac{\ln 2}{|k|}$ ; they do not depend on the starting amount.
To find $k$ from two values, divide first: $\frac{y_1}{y_0} = e^{kt_1}$ , then take $\ln$ .
Cooling toward room temperature $T_r$ is exponential in $T - T_r$ , not in $T$ .

dy/dt = ky y = y_0e^kt , the exponential model growth or decay; |k| is the rate as a fraction of the amount, per unit time e^(kt_1) = (y_1)/(y_0) , so k = 1/(t_1) ln (y_1)/(y_0)

exponential model: A quantity whose rate of change is proportional to its size, $\frac{dy}{dt} = ky$ , so that $y = y_0e^{kt}$ with $y_0 = y(0)$ . The constant $k$ is the growth constant ( $k > 0$ ) or decay constant ( $k < 0$ ): the rate of change as a fraction of the current amount, per unit of time.

7.9 Logistic Models

A logistic model, dP/dt = kP(K − P) , has a rate jointly proportional to the quantity and to the room left. Any positive starting value approaches the carrying capacity K (from below it rises, from above it falls), and growth is fastest at P = K/2 , the point of inflection of P(t) . All of this is read from the equation, without solving it.

Logistic: $\frac{dP}{dt} = kP(K - P)$ or $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ . The carrying capacity $K$ is the positive value of $P$ that makes the rate 0.
For $0 < P(0) < K$ , $P$ increases toward $K$ ; for $P(0) > K$ , $P$ decreases toward $K$ . Either way $\lim_{t \to \infty} P(t) = K$ .
Growth is fastest at $P = \frac{K}{2}$ : below it the graph of $P$ is concave up, above it concave down.
Read these from the equation; no formula for $P(t)$ is needed.

dP/dt = kP(K − P) , a logistic differential equation dP/dt ≈ kKP : growth is nearly exponential growth slows and P levels off: lim as t → ∞, P(t) = K , the carrying capacity it is largest at P = K/2 , where the graph of P(t) has a point of inflection

logistic model: A differential equation in which the rate of change is jointly proportional to the quantity and to the room left below a limit $K$ : $\frac{dP}{dt} = kP(K - P)$ , or $rP\left(1 - \frac{P}{K}\right)$ . Positive solutions approach $K$ , and growth is fastest at $P = \frac{K}{2}$ .
carrying capacity: In a logistic model, the positive value $K$ at which the rate of change is 0; for any positive starting value, $\lim_{t \to \infty} P(t) = K$ . Read it from $\frac{dP}{dt} = kP(K - P)$ by setting the rate to 0.