Unit 07: Differential Equations Flashcards

(33 cards)

1
Q

Differential Equation

A

An equation that relates a function with its derivative, written in the form dy/dx = f(x, y).

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2
Q

Slope Field

A

A graphical tool used to visualize solutions of a first-order differential equation consisting of a grid of short line segments representing slopes.

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3
Q

Slope (at a point)

A

The value of the derivative dy/dx at a specific point, representing the rate of change of the function at that point.

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4
Q

Initial Condition

A

A specific point (x₀, y₀) that the solution curve must pass through; used to find a particular solution.

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5
Q

Particular Solution

A

A single function that satisfies the differential equation and passes through a specific given initial condition.

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6
Q

Constant of Integration (C)

A

The constant added to an antiderivative to represent the family of all possible solutions (the general solution).

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7
Q

Undefined Slope

A

Occurs when the DE has a zero in the denominator; graphically represented by no slope drawn or a vertical asymptote.

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8
Q

Separation of Variables

A

A method to solve DEs by algebraically rearranging terms so that all y/dy terms are on one side and all x/dx terms are on the other.

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9
Q

Leibniz Notation

A

The notation (dy/dx) that treats the derivative as a ratio of differentials, essential for the separation process.

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10
Q

General Solution

A

The solution to a DE that includes an arbitrary constant C, representing a family of functions.

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11
Q

Explicit Form

A

A solution written in the form y = f(x), where the dependent variable y is isolated on one side.

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12
Q

Growth/Decay Differential Equation

A

An equation where the rate of change is proportional to the variable itself: dy/dt = ky.

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13
Q

Exponential Growth and Decay Formula

A

The general solution to the growth/decay DE: y = Ce^(kt).

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14
Q

Initial Amount (C)

A

The quantity present at time t = 0; the starting value.

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15
Q

Growth or Decay Rate (k)

A

A constant representing the proportionality factor. k > 0 means growth; k < 0 means decay.

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16
Q

Half-life

A

The amount of time (t) required for a quantity to fall to half of its initial value (y = 1/2 C).

17
Q

Constructing a Slope Field (Steps)

A
  1. Select coordinates (x,y). 2. Substitute values into DE to calculate slope. 3. Draw a small line segment with that slope.
18
Q

Relationship between dy/dx and Graph Shape

A

> 0: segment slants upward; < 0: segment slants downward; = 0: segment is horizontal.

19
Q

Finding a Particular Solution Graphically

A

Start at the initial condition point and ‘follow the flow’ of the line segments in the slope field.

20
Q

Connection with Integrals

A

Solving a DE algebraically involves the integral: ∫(dy/dx) dx = ∫f(x) dx = F(x) + C.

21
Q

Euler’s Method

A

A numerical technique used to approximate a particular solution to a differential equation.

22
Q

Euler’s Method Formula (x-value)

A

xₙ = xₙ₋₁ + h

23
Q

Euler’s Method Formula (y-value)

A

yₙ = yₙ₋₁ + h · F(xₙ₋₁, yₙ₋₁)

24
Q

Step Size (h)

A

A small distance moved along the x-axis in Euler’s Method; smaller h leads to more accurate approximations.

25
The Strategy for Separation (Steps)
1. Rewrite notation (y' to dy/dx). 2. Separate terms. 3. Integrate both sides. 4. Solve for y (isolate).
26
Handling the Constant C in Separation
Constants from both sides are combined into a single '+C' on the x-side of the equation.
27
Integration Techniques in Separation
Solving the x-side often requires advanced methods like u-substitution.
28
Logarithmic Solutions
If integration results in ln|y| = f(x) + C, you must exponentiate both sides (base e) to solve for y.
29
Isolating y: y² = f(x) + C
y = ±√(f(x) + C)
30
Isolating y: y³ = f(x) + C
y = ∛(f(x) + C)
31
Deriving Growth Formula: Exponentiation Step
e^(ln|y|) = e^(kt+C₁) becomes y = e^(kt) · e^(C₁).
32
Properties of Exponents in Growth/Decay
The term e^C is just a constant and is replaced by the coefficient C in the final equation y = Ce^(kt).
33
Finding C and k with Two Data Points
1. Create two equations: y₁ = Ce^(kt₁) and y₂ = Ce^(kt₂). 2. Isolate C in one. 3. Substitute into the second to solve for k. 4. Solve for C.