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Flashcards · AP® Calculus AB · Unit 7

Unit 7: Differential Equations flashcards

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7.1 Modeling Situations with Differential Equations

What is a differential equation, and what is its solution?
An equation linking a function, its independent variable and one or more of its derivatives, such as dP/dt = 0.03P. Its solution is a function, not a number.
How do you translate "A changes at a rate proportional to A" and "inversely proportional to A"?
dA/dt = kA and dA/dt = k/A. "Inversely" means k/A, not a negative sign.
How do you write "jointly proportional to A and (M − A)" and "proportional to the difference between A and R"?
dA/dt = kA(M − A) and dA/dt = k(A − R).
The mass m of a snowball shrinks at a rate proportional to its surface area S. Write the differential equation.
dm/dt = −kS with k > 0. The minus sign shows the mass is decreasing; avoid a double negative by stating k > 0.
dy/dt = −k(y − 15) with k > 0, and y = 35 when dy/dt = −4. Find k, then dy/dt when y = 25.
−4 = −k(20), so k = 0.2. When y = 25, dy/dt = −0.2 × 10 = −2.
What are the units of k in dA/dt = kA with t in hours?
Per hour. In general, units of k = (units of dA/dt) ÷ (units of what k multiplies).
What mistakes are common when writing differential equations from words?
Writing the quantity, not its rate (P = kP); a double negative with k negative; "proportional to time" (kt) versus "to the amount" (kA); "cooling at 3 °C per minute" means dT/dt = −3.

7.2 Verifying Solutions for Differential Equations

How do you verify that a function is a solution of a differential equation?
Find y′ (and y″ if needed), substitute into the left side and the right side separately, simplify each, and show they are identical for every input in the interval.
Does agreement at one x value prove a function is a solution?
No. Agreement at one point proves nothing, but disagreement at one point does disprove it.
Is y = 7e^(−2x) a solution of dy/dx = −2y?
Yes. dy/dx = −14e^(−2x), and −2y = −14e^(−2x), so both sides agree for all x.
Find r so that y = e^(rx) solves y″ = 9y.
y″ = r²e^(rx) = 9e^(rx) for all x needs r² = 9, so r = 3 or r = −3.
What is the family of solutions of dy/dx = ky?
y = Ce^(kx), since y′ = kCe^(kx) = ky. A differential equation usually has infinitely many solutions.
When is a constant function y = c a solution?
When substituting y = c makes the right side 0 for every x, since its derivative is 0.

7.3 Sketching Slope Fields

What is a slope field?
For dy/dx = f(x, y), a grid of short segments where each segment at (x, y) has slope f(x, y). Each segment is a piece of the tangent line to the solution through that point.
How can you tell from a slope field whether dy/dx depends on x only or on y only?
Identical columns: dy/dx depends on x only. Identical rows: dy/dx depends on y only. Neither repeats: it depends on both.
For dy/dx = xy − 2, what is the slope of the segment at (2, 3)?
2 × 3 − 2 = 4: a steep rise. Take x first and y second.
For dy/dx = 3 − y, which segments are horizontal?
All segments along the row y = 3. Along y = 5 every segment has slope −2, so each row is identical.
How should slope-field segments be drawn?
Short, equal in length and centred on the grid points. Only the tilt carries information. Draw a horizontal segment where the slope is 0; never leave it blank.
Why is matching slope fields on zero slopes alone risky?
Different equations can share flat rows: 1 − y² and y² − 1 both have zero slope at y = ±1 but opposite signs elsewhere. Check the sign at a test point.

7.4 Reasoning Using Slope Fields

How do you sketch a particular solution on a slope field?
Start at the given point, follow the segments to the right and to the left, keep the curve smooth and tangent to the segments, and stop at the edge of the field.
What is a constant solution on a slope field, and how do nearby solutions behave?
A horizontal line y = c along which every segment is flat. Nearby solutions level off towards it or move away, but do not cross it.
Find the constant solutions of dy/dx = y(y − 4). Is the solution with y(0) = 2 increasing or decreasing?
y = 0 and y = 4. At y = 2, dy/dx = 2(−2) = −4, so it is decreasing and falls towards y = 0.
For dy/dx = x + y, find d²y/dx² and its sign at (1, −3).
d²y/dx² = 1 + dy/dx. At (1, −3), dy/dx = −2, so d²y/dx² = −1: concave down.
Is y = 2x a solution of dy/dx = 2x − y?
No. Its slope is 2, but on that line the equation gives 2x − 2x = 0. A sloping line of flat segments is not a solution.
How do you judge a tangent-line estimate for a solution of a differential equation?
Use the sign of d²y/dx² at the point: concave up gives an underestimate, concave down an overestimate.

7.6 Finding General Solutions Using Separation of Variables

When is a differential equation separable?
When dy/dx = g(x)·h(y). Try factoring (xy + 2x = x(y + 2)) and exponent laws (e^(x + y) = eˣeʸ) first. Sums like x + y or cos(x + y) are not separable.
What is the method of separation of variables?
Write (1/h(y)) dy = g(x) dx, integrate both sides, add one constant C, then solve for y (or leave it implicit) and check by differentiating.
Find the general solution of dy/dx = 2x/y.
y dy = 2x dx, so y²/2 = x² + C, or y² = 2x² + K.
Find the general solution of dy/dx = y sec²x.
(1/y) dy = sec²x dx, so ln|y| = tan x + C, giving y = Ae^(tan x) with A any real constant.
How do you handle ln|y| = G(x) + C?
|y| = e^C·e^(G(x)), so y = Ae^(G(x)). The constant multiplies; writing e^(G(x)) + C is wrong.
What must you check separately when separating variables?
You divide by h(y), so constant solutions where h(y) = 0 (such as y = 0) must be checked; often A = 0 covers them.
What mistakes are common with separation of variables?
No constant; adding C after rearranging; a logarithm for every fraction (∫ 1/y² dy = −1/y); integrating y with respect to x without separating; using two constants where one will do.

7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables

How do you find a particular solution from an initial condition?
Separate, integrate both sides, add one C, substitute (a, y₀) into the integrated equation, solve for C and y, choose the sign that matches y₀, state the domain, and check.
Solve dy/dx = 6x² with y(1) = 4.
y = 2x³ + C; 4 = 2 + C, so C = 2 and y = 2x³ + 2.
Solve dy/dx = x/y with y(0) = −2.
y dy = x dx gives y² = x² + C; at (0, −2), C = 4. Since y₀ is negative, y = −√(x² + 4) for all real x.
What is the domain of a particular solution?
The largest open interval containing the initial x-value on which the formula is defined and satisfies the equation. For dy/dx = y², y(0) = 1, y = 1/(1 − x) has domain x less than 1.
How can you write a particular solution of dy/dx = f(x), y(a) = y₀ without an antiderivative?
y = y₀ + ∫ (a to x) f(t) dt. Use a calculator for values if no antiderivative is available.
What mistakes are common with particular solutions?
Putting (a, y₀) into dy/dx (that only gives a slope); assuming C = y₀; adding C after rearranging; leaving ± in the answer; giving a domain with a gap.

7.8 Exponential Models with Differential Equations

What is the solution of dy/dt = ky with y(0) = y₀?
y = y₀e^(kt). k > 0 means growth, k negative means decay, and k has units of per unit time.
How do you find k from a doubling time or a half-life?
Doubling time D: k = ln 2/D. Half-life H: k = −ln 2/H, and y = y₀·2^(−t/H).
How do you find k from two values y(0) = y₀ and y(T) = y₁?
y₁ = y₀e^(kT), so k = ln(y₁/y₀)/T. A quantity that triples every 5 years has k = ln 3/5 ≈ 0.220 per year.
dQ/dt = −0.1Q with t in days. Find the half-life.
ln 2/0.1 ≈ 6.931 days.
y = 3e^(kt) with k = (ln 2)/2. Find y(4) without a calculator.
12. The doubling time is 2, so y doubles twice: 3 → 6 → 12.
What does k mean in dy/dt = ky?
The relative rate: (dy/dt)/y = k, a fraction of the current amount per unit time. Over any time step Δt, y is multiplied by e^(kΔt).
What is the solution of dy/dt = k(y − M)?
y − M = (y₀ − M)e^(kt): the difference from M grows or decays exponentially.
What mistakes are common with exponential models?
Writing y = e^(kt) + C instead of Ce^(kt); treating k as a fixed amount per unit time; using rate × time for a non-constant rate; rounding k early.

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