Difficulty: Intermediate | Prerequisites: Basic integration (power rule, trig integrals, exponentials)
This section covers the reverse of differentiation: finding antiderivatives, solving initial value problems (where you also pin down the constant of integration), and handling separable differential equations. It also includes two applied settings, projectile motion and logistic growth, that tie the algebra to something physical.
Antiderivative
A function F(x) whose derivative is f(x). Also called an indefinite integral. Always includes + C because the derivative of a constant is zero.
Initial value problem (IVP)
A differential equation paired with a condition like y(0) = 5 that fixes the constant of integration. The result is one specific solution, not a family.
Separable differential equation
A first-order ODE that can be written as g(y) dy = h(t) dt, so that all y terms are on one side and all t terms on the other. You then integrate both sides independently.
Think of it as physically pulling y and t apart before integrating.
General solution
The family of all antiderivatives, written with + C.
Particular solution
The single member of that family picked out by an initial condition.
Logistic growth model
A population model where growth slows as the population approaches a carrying capacity L. The solution has the form P(t) = L / (1 + Ce^(-kt)), where C is determined by initial conditions.
In simple terms, the population grows fast at first, then levels off at L.
Given y' = 12t^5 - 12t^2 + 9t - 18t^(-4), find y.
Integrate each term using the power rule:
12t^5 integrates to 2t^6.
-12t^2 integrates to -4t^3.
9t integrates to (9/2)t^2.
-18t^(-4) integrates to 6t^(-3).
General solution: y = 2t^6 - 4t^3 + (9/2)t^2 + 6t^(-3) + C.
Given y'' = 20e^(4t) + sin(5t), find y.
First integration (to get y'):
20e^(4t) integrates to 5e^(4t).
sin(5t) integrates to -cos(5t)/5.
y' = 5e^(4t) - cos(5t)/5 + C_1.
Second integration (to get y):
5e^(4t) integrates to (5/4)e^(4t).
-cos(5t)/5 integrates to -sin(5t)/25.
y = (5/4)e^(4t) - sin(5t)/25 + C_1 t + C_2.
Two constants appear because you integrated twice.
Given y' = 2t + sin(t), y(0) = -2. Find y.
Integrate: y = t^2 - cos(t) + C.
Apply the initial condition: -2 = 0 - cos(0) + C = -1 + C, so C = -1.
Particular solution: y = t^2 - cos(t) - 1.
A ball is launched upward with initial velocity 49 m/s from a height of 35 m. Acceleration a(t) = -9.8 m/s^2.
Integrate acceleration to get velocity: v(t) = -9.8t + 49.
Integrate velocity to get position: s(t) = -4.9t^2 + 49t + 35.
The ball reaches maximum height when v(t) = 0: t = 49/9.8 = 5 seconds.
Maximum height: s(5) = -4.9(25) + 49(5) + 35 = -122.5 + 245 + 35 = 157.5 m.
Given y' = y(4t^3 + 5t^4), y(0) = 9.
Separate: (1/y) dy = (4t^3 + 5t^4) dt.
Integrate both sides: ln|y| = t^4 + t^5 + C.
Apply initial condition: ln(9) = 0 + 0 + C, so C = ln(9).
Exponentiate: y = e^(ln 9) times e^(t^4 + t^5) = 9e^(t^4 + t^5).
Given 2y y' = 15t^2, y(0) = 9.
Rewrite as 2y dy = 15t^2 dt (already separated).
Integrate: y^2 = (5/2)t^3 + C (since integral of 2y dy = y^2).
Wait, more carefully: integral of 2y dy = y^2, and integral of 15t^2 dt = 5t^3. So y^2 = 5t^3 + C.
Apply initial condition: 81 = 0 + C, so C = 81.
y = sqrt(5t^3 + 81).
A logistic model P(t) = L / (1 + Ce^(-kt)).
Given information: the carrying capacity is 350, and the initial population is 70.
Find C: at t = 0, P(0) = 350 / (1 + C) = 70. So 1 + C = 5, giving C = 4.
Solution: P(t) = 350 / (1 + 4e^(-0.08t)).
The limiting value as t goes to infinity is L = 350, since the exponential term vanishes.
Power rule for integration:
\int t^n \, dt = \frac{t^{n+1}}{n+1} + C, \quad n \neq -1Separation of variables procedure:
Rewrite dy/dt = f(t) g(y) as (1/g(y)) dy = f(t) dt.
Integrate both sides.
Solve for y if possible.
Apply the initial condition to find C.
Logistic solution:
P(t) = \frac{L}{1 + C e^{-kt}}where L is the carrying capacity, k is the growth rate, and C = (L - P_0) / P_0.
Forgetting + C on indefinite integrals. Every antiderivative has a family of solutions; the constant is not optional until an initial condition pins it down.
When integrating y'' twice to get y, students sometimes use only one constant. Each integration introduces its own constant: C_1 and C_2.
In separable ODEs, students sometimes leave the answer as ln|y| = ... without exponentiating. Unless the problem says otherwise, solve explicitly for y.
In projectile motion, students sometimes forget the initial height. If the object starts at height h_0, the position function is s(t) = -(g/2)t^2 + v_0 t + h_0, not just -(g/2)t^2 + v_0 t.
Straight antiderivative problems ("find y given y'") are quick marks if your integration is accurate. Watch for negative exponents and trig integrals.
IVPs are extremely common. The pattern is always: integrate, apply condition, solve for C.
Separable ODEs on the exam follow a predictable template: separate, integrate, apply initial condition, simplify. Practise the algebra of exponentiating ln|y|.
Expect one projectile motion problem and one logistic growth problem. For logistic growth, be ready to find C from P(0) and to state the long-run limit (always L).
Fill in the blank: The antiderivative of t^5 is ___ + C. (t^6/6.)
True or false: If y' = sin(t) and y(0) = 3, then y = -cos(t) + 4. (True, since y = -cos(t) + C and -cos(0) + C = -1 + C = 3 gives C = 4.)
Fill in the blank: In a separable ODE, the first step is to move all y terms to ___ side and all t terms to the other. (One.)
True or false: The carrying capacity in a logistic model is the value the population approaches as t goes to infinity. (True.)
True or false: Integrating y'' twice introduces only one constant of integration. (False, it introduces two: C_1 and C_2.)
Q: Given y' = 12t^5 - 12t^2 + 9t - 18t^(-4), find the general antiderivative.
A: y = 2t^6 - 4t^3 + (9/2)t^2 + 6t^(-3) + C.
Q: Solve the IVP y' = 2t + sin(t), y(0) = -2.
A: Integrate to get y = t^2 - cos(t) + C. Apply y(0) = -2: -2 = 0 - 1 + C, so C = -1. Solution: y = t^2 - cos(t) - 1.
Q: Solve y' = y(4t^3 + 5t^4), y(0) = 9.
A: Separate and integrate: ln|y| = t^4 + t^5 + C. At t = 0: C = ln 9. Solution: y = 9e^(t^4 + t^5).
Q: A ball is launched upward at 49 m/s from 35 m. When does it reach maximum height, and what is that height?
A: v(t) = -9.8t + 49 = 0 at t = 5 s. s(5) = -4.9(25) + 49(5) + 35 = 157.5 m.
Q: In a logistic model with L = 350 and P(0) = 70, what is C?
A: 350/(1 + C) = 70, so 1 + C = 5, giving C = 4.
Antiderivatives are the foundation of the Fundamental Theorem of Calculus, covered earlier in M 2414. Separable ODEs reappear in more advanced differential equations courses, where you will also meet non-separable types. The logistic model connects to population biology and is a standard example in applied maths courses. The series techniques from the earlier notes (Taylor, Maclaurin) can also be used to solve ODEs via power series methods in more advanced study.
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Antiderivative, indefinite integral, initial value problem, IVP, separable differential equation, separation of variables, general solution, particular solution, constant of integration, projectile motion, kinematics, logistic growth, carrying capacity, logistic equation, M 2414, integral calculus, UT Austin