Unit 7 Beta

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 <math class="calc-m"><semantics><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>=</mo><mn>0.3</mn><mi>y</mi></mrow><annotation encoding="application/x-tex">\frac{dy}{dt} = 0.3y</annotation></semantics></math>. It says how fast a quantity changes, often in terms of the quantity itself.
proportional to: "<math class="calc-m"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math> is proportional to <math class="calc-m"><semantics><mrow><mi>B</mi></mrow><annotation encoding="application/x-tex">B</annotation></semantics></math>" means <math class="calc-m"><semantics><mrow><mi>A</mi><mo>=</mo><mi>k</mi><mi>B</mi></mrow><annotation encoding="application/x-tex">A = kB</annotation></semantics></math> for a constant <math class="calc-m"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math>, the constant of proportionality. "Inversely proportional to <math class="calc-m"><semantics><mrow><mi>B</mi></mrow><annotation encoding="application/x-tex">B</annotation></semantics></math>" means <math class="calc-m"><semantics><mrow><mi>A</mi><mo>=</mo><mfrac><mi>k</mi><mi>B</mi></mfrac></mrow><annotation encoding="application/x-tex">A = \frac{k}{B}</annotation></semantics></math>.

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 <math class="calc-m"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math> 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 <math class="calc-m"><semantics><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mi>F</mi><mo stretchy="false">(</mo><mi>x</mi><mo>,</mo><mi>y</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\frac{dy}{dx} = F(x, y)</annotation></semantics></math>: at each chosen point <math class="calc-m"><semantics><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>,</mo><mi>y</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(x, y)</annotation></semantics></math>, a short segment with slope <math class="calc-m"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>x</mi><mo>,</mo><mi>y</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(x, y)</annotation></semantics></math>. 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 <math class="calc-m"><semantics><mrow><mi>y</mi><mo>=</mo><mi>c</mi></mrow><annotation encoding="application/x-tex">y = c</annotation></semantics></math> that solves a differential equation: <math class="calc-m"><semantics><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\frac{dy}{dx} = 0</annotation></semantics></math> at every point of the horizontal line <math class="calc-m"><semantics><mrow><mi>y</mi><mo>=</mo><mi>c</mi></mrow><annotation encoding="application/x-tex">y = c</annotation></semantics></math>. In a slope field it is a row of horizontal segments.

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 <math class="calc-m"><semantics><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mi>h</mi><mo stretchy="false">(</mo><mi>y</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\frac{dy}{dx} = g(x)h(y)</annotation></semantics></math>: move every <math class="calc-m"><semantics><mrow><mi>y</mi></mrow><annotation encoding="application/x-tex">y</annotation></semantics></math> to the side with <math class="calc-m"><semantics><mrow><mi>d</mi><mi>y</mi></mrow><annotation encoding="application/x-tex">dy</annotation></semantics></math> and every <math class="calc-m"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math> to the side with <math class="calc-m"><semantics><mrow><mi>d</mi><mi>x</mi></mrow><annotation encoding="application/x-tex">dx</annotation></semantics></math>, <math class="calc-m"><semantics><mrow><mfrac><mn>1</mn><mrow><mi>h</mi><mo stretchy="false">(</mo><mi>y</mi><mo stretchy="false">)</mo></mrow></mfrac><mspace width="0.1667em"></mspace><mi>d</mi><mi>y</mi><mo>=</mo><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mspace width="0.1667em"></mspace><mi>d</mi><mi>x</mi></mrow><annotation encoding="application/x-tex">\frac{1}{h(y)}\,dy = g(x)\,dx</annotation></semantics></math>, then integrate both sides and add one constant <math class="calc-m"><semantics><mrow><mi>C</mi></mrow><annotation encoding="application/x-tex">C</annotation></semantics></math>. The result, a family with <math class="calc-m"><semantics><mrow><mi>C</mi></mrow><annotation encoding="application/x-tex">C</annotation></semantics></math> 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 <math class="calc-m"><semantics><mrow><mi>y</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">y(0) = 1</annotation></semantics></math> (a point the curve must pass through), together with its domain: the largest open interval containing the initial <math class="calc-m"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math> 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, <math class="calc-m"><semantics><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>=</mo><mi>k</mi><mi>y</mi></mrow><annotation encoding="application/x-tex">\frac{dy}{dt} = ky</annotation></semantics></math>, so that <math class="calc-m"><semantics><mrow><mi>y</mi><mo>=</mo><msub><mi>y</mi><mn>0</mn></msub><msup><mi>e</mi><mrow><mi>k</mi><mi>t</mi></mrow></msup></mrow><annotation encoding="application/x-tex">y = y_0e^{kt}</annotation></semantics></math> with <math class="calc-m"><semantics><mrow><msub><mi>y</mi><mn>0</mn></msub><mo>=</mo><mi>y</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">y_0 = y(0)</annotation></semantics></math>. The constant <math class="calc-m"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math> is the growth constant (<math class="calc-m"><semantics><mrow><mi>k</mi><mo>></mo><mn>0</mn></mrow><annotation encoding="application/x-tex">k > 0</annotation></semantics></math>) or decay constant (<math class="calc-m"><semantics><mrow><mi>k</mi><mo><</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">k < 0</annotation></semantics></math>): the rate of change as a fraction of the current amount, per unit of time.