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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Each solution shows the separation, the integration with its constant, and the use of the initial condition. If your answer differs, check that the constant was added at the integration step and that the correct sign of a square root was chosen.
Standards: AP Calculus AB Unit 7 · Level: Grade 12 AP Calculus AB
Problem 1 Verifying an exponential solution
Differentiate: y′ = 3·2e2x = 6e2x. The right side is 2y = 2·3e2x = 6e2x. Both sides match, so the function is a solution.
Problem 2 General and particular solutions
The derivative of x2 + C is 2x for any constant. For y(1) = 5: 1 + C = 5, so C = 4. The particular solution is y = x2 + 4.
Problem 3 Verifying a second-order solution
y′ = 10 cos 2x + 1 and y″ = −20 sin 2x. Then y″ + 4y = −20 sin 2x + 20 sin 2x + 4x = 4x. The equation is satisfied.
Problem 4 Computing slopes from an equation
Substitute each point: at (1, 2) the slope is 1 + 2 = 3; at (−2, 1) it is −2 + 1 = −1; at (0, 0) it is 0.
Problem 5 Reading a slope field
The slope is zero when x − y = 0, so along the line y = x. At (0, 1), dydx = 0 − 1 = −1 < 0, so the solution is decreasing there.
Problem 6 Separable equation with a power
Separate: dy/y = 4x3 dx. Integrate: ln|y| = x4 + C, so y = Aex4. With y(0) = 5, A = 5. The solution is y = 5ex4.
Problem 7 Separable equation with a quotient
Separate: y dy = x dx. Integrate: y2/2 = x2/2 + C, so y2 = x2 + K. With (2, 3): 9 = 4 + K, so K = 5. Since y(2) > 0, the solution is y = √(x2 + 5).
Problem 8 Population growth model
The solution of dP/dt = kP is P = P0ekt. So P(t) = 400e0.05t and P(10) = 400e0.5 ≈ 659.5 bacteria.
Problem 9 Half-life
The decay constant is k = −(ln 2)/8 ≈ −0.0866. The amount is y = 90e−0.0866t, equivalently 90(1/2)t/8. At t = 20: 90(1/2)2.5 = 90/5.657 ≈ 15.9 mg.
Problem 10 Equation with an exponential
Write ex − y = ex/ey. Separate: ey dy = ex dx. Integrate: ey = ex + C. At (0, ln 3): 3 = 1 + C, so C = 2. Therefore y = ln(ex + 2).
Problem 11 Equation with a trigonometric factor
Separate: dy/y = sin x dx. Integrate: ln|y| = −cos x + C, so y = Ae−cos x. At x = 0: 3 = Ae−1, so A = 3e. The solution is y = 3e1 − cos x.
Problem 12 Newton’s law of cooling
Separate: dT/(T − 70) = −0.1 dt. Integrate: ln|T − 70| = −0.1t + C, so T − 70 = Ae−0.1t. With T(0) = 190, A = 120. So T = 70 + 120e−0.1t, and T(10) = 70 + 120e−1 ≈ 114.1°F.
Problem 13 Equilibrium and long-term behavior
Slopes are zero when y(4 − y) = 0, so the equilibria are y = 0 and y = 4. Starting at (0, 1), the slope is positive and the curve rises toward y = 4, so the limit is 4. Differentiating, d2ydx2 = (4 − 2y)dydx, which is positive when 0 < y < 2. The solution is concave up until y reaches 2.
Problem 14 Separable equation with a cube root
Separate: y2 dy = (x + 1)2 dx. Integrate: y3/3 = (x + 1)3/3 + C, so y3 = (x + 1)3 + K. At (0, 2): 8 = 1 + K, so K = 7. The solution is y = ∛((x + 1)3 + 7).
Problem 15 Fish population
With P = 2,400ekt, 3,100 = 2,400e3k, so k = ln(31/24)/3 ≈ 0.0853 per year. Then P(10) = 2,400e0.853 ≈ 5,633 fish.
Problem 16 Salt in a tank
Rewrite: dS/dt = (300 − S)/50. Separate: dS/(300 − S) = dt/50. Integrate: −ln|300 − S| = t/50 + C, so 300 − S = Ae−t/50. With S(0) = 0, A = 300. So S = 300(1 − e−t/50). At t = 50: 300(1 − e−1) ≈ 189.6 lb. As t → ∞ the amount approaches 300 lb.
Problem 17 Draining tank
Separate: h−1/2 dh = −0.3 dt. Integrate: 2√h = −0.3t + C. At t = 0: 2(4) = 8 = C. So √h = 4 − 0.15t and h = (4 − 0.15t)2. The tank is empty when 4 − 0.15t = 0, so t = 4/0.15 ≈ 26.7 minutes.
Problem 18 Choosing the sign of a root
Separate: y dy = (x + 2) dx. Integrate: y2/2 = x2/2 + 2x + C. At (0, −1): 1/2 = C. Multiply by 2: y2 = x2 + 4x + 1. Since y(0) = −1 is negative, take the negative root: y = −√(x2 + 4x + 1).
Problem 19 Equilibrium and exponential behavior
The equilibrium is y = 3, where the slope is zero. For y ≠ 3, separate: dy/(y − 3) = x dx. Integrate: ln|y − 3| = x2/2 + C, so y − 3 = Aex2/2. At (0, 5): A = 2. So y = 3 + 2ex2/2 and y(2) = 3 + 2e2 ≈ 17.8.
Problem 20 Time to cool
Separate and integrate: T − 22 = Ae−0.08t. With T(0) = 96, A = 74. So T = 22 + 74e−0.08t. Set T = 50: 28 = 74e−0.08t, so t = ln(74/28)/0.08 ≈ 12.1 minutes. Because e−0.08t is never zero, T approaches but never reaches 22°F.
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