Antiderivatives and Initial Value Problems
Quick Reference
| Textbook | J. David Logan, A First Course in Differential Equations, 3rd ed. (Springer, 2015) |
| Chapter.Section.Subsection | 1.2 Antiderivatives |
| Pages | 16 to 21 |
| Reads on | Theorem 1.12 (Fundamental Theorem of Calculus), Examples 1.9 through 1.11, Eqs. (1.7) through (1.9) |
The page range above is verified against the author’s complete PDF of the 3rd edition hosted at the University of Nebraska-Lincoln. Bring a copy of Logan to this topic; the section and pages are the exact reading for this skill.
Try This First
A car has velocity $v(t) = 6t$ meters per second at every time $t \geq 0$, and at the start ($t = 0$) it sits at position $x = 4$ meters.
Before any formula, answer two questions on paper.
- The position $x(t)$ is the quantity whose rate of change is $6t$. Write down a function whose derivative is $6t$. (Try $x(t) = 3t^2$ and differentiate it to check.)
- Your function from step 1, does it give $x = 4$ at $t = 0$? If not, what number would you add to fix the starting value, without changing the rate of change?
Check your thinking
- The derivative of $3t^2$ is $6t$, so $x(t) = 3t^2$ has the right rate of change. So does $3t^2 + 1$, and $3t^2 - 7$, and $3t^2$ plus any constant. Adding a constant does not change the derivative, because the derivative of a constant is zero.
- At $t = 0$, the bare function $3t^2$ gives $0$, not $4$. Adding $4$ fixes the start: $x(t) = 3t^2 + 4$ gives $x(0) = 4$ and still has rate of change $6t$.
You just solved an initial value problem by hand. The rate law $x' = 6t$ has a whole family of position functions $3t^2 + C$, and the starting value $x(0) = 4$ selected the one with $C = 4$.
Intuition Before Symbols
A differential equation of the form $x' = g(t)$ hands you the rate at which a quantity changes and asks you to recover the quantity itself. The rate $g(t)$ depends only on the clock time $t$, not on the current amount $x$, so the recovery is pure undoing of a derivative: antidifferentiation.
Undoing a derivative cannot give a single answer. Many functions share the same derivative, because adding a constant leaves the derivative untouched. So the rate law alone produces a family of functions, one for each constant. To single out one member of the family you need one more fact: where the quantity starts. That extra fact is the initial condition, and the rate law together with the initial condition is an initial value problem.
Two representations sit side by side here. Symbolically, $x(t) = \int g(t)\,dt + C$ is a formula with a free constant. Geometrically, the formula is a stack of parallel curves, each a vertical shift of the others, all rising at the same rate $g(t)$ at any given time. The initial condition is a single point the curve must pass through, and exactly one curve in the stack passes through it.
Prerequisite Hub
Builds On
| Skill | Course | Why it is needed here |
|---|---|---|
Antiderivatives and the Constant of Integration (math347-w0-antiderivatives-review) |
MATH347 | The general solution of $x' = g(t)$ is an antiderivative with $+C$ |
Introduction to Differential Equations (de-introduction) |
MATH347 | Order, solution, and the meaning of an initial-value problem |
Antiderivative concept (antiderivative-concept) |
MATH162 | Reversing differentiation to find an indefinite integral |
Fundamental Theorem of Calculus, Part 2 (ftc-part2) |
MATH162 | Writing a solution as a definite integral with a variable upper limit |
Initial value problems (initial-value-problems) |
MATH161 | Using one condition to determine an unknown constant |
Unlocks
| Skill | Course | What this skill provides for it |
|---|---|---|
Separable Differential Equations (separable-de) |
MATH347 | The integrate-then-apply-the-condition habit carries directly into separation of variables |
This skill is the bridge between the calculus review of antiderivatives and the first genuine solution technique. The pattern learned here, find the general solution by integrating then pin the constant with the initial condition, is the skeleton of every method that follows.
Definitions From the Textbook
Solving $x' = g(t)$ by antidifferentiation
To solve $x' = g(t)$ for the unknown $x = x(t)$, where $g$ is continuous, write \[ x(t) = \int g(t)\,dt + C, \] an arbitrary constant of integration. This one-parameter family of integral curves is the general solution, unique only up to the additive constant $C$.
Source: Logan, 3rd ed., §1.2, p.16, Eqs. (1.7) through (1.8).
Initial value problem (IVP)
Imposing an initial condition of the form $x(t_0) = x_0$ selects the integral curve passing through the point $(t_0, x_0)$, which determines the constant $C$. The differential equation together with the initial condition is an initial value problem; its solution is the resulting particular solution.
Source: Logan, 3rd ed., §1.2, pp.16-17, Eq. (1.9), Example 1.9.
Solutions defined by an integral
When the antiderivative has no elementary closed form (for example $x' = e^{-t^2}$), the solution must be written using the Fundamental Theorem of Calculus as a definite integral with a variable upper limit, \[ x(t) = \int_{t_0}^{t} g(s)\,ds + x_0; \] some such integrals define special functions.
Source: Logan, 3rd ed., §1.2, pp.17-18, Example 1.11.
Key Theorem
Fundamental Theorem of Calculus (variable upper limit)
If $g(t)$ is a continuous function, the derivative of an integral with variable upper limit is \[ \frac{d}{dt}\int_a^t g(s)\,ds = g(t), \] where the lower limit $a$ is any number.
Source: Logan, 3rd ed., Theorem 1.12, p.18.
This theorem is the reason the definite-integral formula above is a genuine solution. Differentiating $\displaystyle x(t) = \int_{t_0}^{t} g(s)\,ds + x_0$ with respect to $t$ returns $g(t)$, so the rate law $x' = g(t)$ holds, and at $t = t_0$ the integral collapses to zero, so $x(t_0) = x_0$. Both halves of the IVP are satisfied at once.
Quick Reference
| Step | What you do | Why |
|---|---|---|
| 1. Integrate | $x(t) = \int g(t)\,dt + C$ | Undo the derivative; the $+C$ is the whole family |
| 2. Apply the condition | substitute $t = t_0$, $x = x_0$ | Turns the family into one curve through $(t_0, x_0)$ |
| 3. Solve for $C$ | algebra on the equation from step 2 | Names the single constant |
| 4. Write the particular solution | put the value of $C$ back into the formula | The one function that satisfies both the equation and the start |
General versus particular. Step 1 gives the general solution (a family with a free $C$). Steps 2 through 4 give the particular solution (one member of the family).
When the integral has no formula. If $\int g(t)\,dt$ cannot be written with elementary functions, skip the indefinite integral and write the answer as a definite integral with variable upper limit: $\displaystyle x(t) = \int_{t_0}^{t} g(s)\,ds + x_0$. The Fundamental Theorem of Calculus guarantees this is correct.
Worked Examples
Every example follows predict-then-check: estimate the shape or sign of the answer first, then compute, then compare.
Example 1: A first initial value problem
Solve the IVP $x' = 4t$, $\;x(0) = 5$.
Predict. The rate $4t$ is zero at $t = 0$ and grows with $t$, so the solution should start flat at height $5$ and curve upward like a parabola.
Solve. Integrate to get the general solution: \[ x(t) = \int 4t\,dt + C = 2t^2 + C. \] Apply the initial condition by substituting $t = 0$, $x = 5$: \[ 5 = 2(0)^2 + C \implies C = 5. \] The particular solution is \[ x(t) = 2t^2 + 5. \]
Check. Differentiate the answer: $x'(t) = 4t$, which matches the equation. Evaluate at the start: $x(0) = 2(0)^2 + 5 = 5$, which matches the condition. The curve is an upward parabola through $(0, 5)$, exactly as predicted.
Example 2: A nonzero start time
Solve the IVP $x' = 3t^2 - 2$, $\;x(1) = 4$.
Predict. The right side is negative for small $t$ (at $t = 1$ it is $3 - 2 = 1$, already positive) and grows as $t$ increases, so the solution is increasing near $t = 1$.
Solve. General solution: \[ x(t) = \int (3t^2 - 2)\,dt + C = t^3 - 2t + C. \] Apply $x(1) = 4$. Substitute $t = 1$, $x = 4$: \[ 4 = (1)^3 - 2(1) + C = 1 - 2 + C = -1 + C \implies C = 5. \] The particular solution is \[ x(t) = t^3 - 2t + 5. \]
Check. Differentiate: $x'(t) = 3t^2 - 2$, matches. Evaluate at the start: $x(1) = 1 - 2 + 5 = 4$, matches. The start time was $t_0 = 1$, not $0$, and the method did not change: integrate first, then substitute the given point.
Example 3: An exponential rate law
Solve the IVP $x' = e^{2t}$, $\;x(0) = \dfrac{1}{2}$.
Predict. The rate $e^{2t}$ is always positive and grows fast, so the solution increases and steepens. At $t = 0$ the rate is $e^0 = 1$, so the curve starts with slope $1$.
Solve. General solution: \[ x(t) = \int e^{2t}\,dt + C = \frac{1}{2}e^{2t} + C. \] Apply $x(0) = \dfrac{1}{2}$: \[ \frac{1}{2} = \frac{1}{2}e^{0} + C = \frac{1}{2} + C \implies C = 0. \] The particular solution is \[ x(t) = \frac{1}{2}e^{2t}. \]
Check. Differentiate: $x'(t) = \dfrac{1}{2}\cdot 2 e^{2t} = e^{2t}$, matches. Evaluate at the start: $x(0) = \dfrac{1}{2}e^0 = \dfrac{1}{2}$, matches. Here the constant came out to exactly zero, a reminder that $C = 0$ is a legitimate value, not a sign that something was skipped.
Example 4: When the antiderivative has no elementary formula
Solve the IVP $x' = e^{-t^2}$, $\;x(0) = 1$.
Predict. The rate $e^{-t^2}$ is always positive and largest at $t = 0$, so the solution increases everywhere, fastest near the start. A formula with elementary functions is not expected, because $e^{-t^2}$ has no elementary antiderivative.
Solve. Attempting $\int e^{-t^2}\,dt$ produces no answer in terms of powers, exponentials, logarithms, or trig functions. Instead, write the solution directly with a definite integral and a variable upper limit, using the Fundamental Theorem of Calculus: \[ x(t) = \int_{0}^{t} e^{-s^2}\,ds + 1. \] The dummy variable $s$ is the integration variable, and $t$ is the variable upper limit. The added $1$ is the initial value $x_0$.
Check. Differentiate with Theorem 1.12: $\displaystyle x'(t) = \frac{d}{dt}\int_0^t e^{-s^2}\,ds + 0 = e^{-t^2}$, matches the equation. Evaluate at the start: $\displaystyle x(0) = \int_0^0 e^{-s^2}\,ds + 1 = 0 + 1 = 1$, matches the condition. This integral is so important in probability and physics that it has its own name (related to the error function). Leaving the answer as a definite integral is the complete and correct solution, not an unfinished one.
Example 5: Reading a unit and interpreting the answer
A tank fills so that water enters at the rate $V'(t) = 2t + 1$ liters per minute, and the tank holds $3$ liters at $t = 0$. Find the volume $V(t)$ and the volume after $4$ minutes.
Predict. Volume only increases (the rate is positive for $t \geq 0$), starting from $3$ liters, so the answer at $t = 4$ should be comfortably above $3$.
Solve. General solution: \[ V(t) = \int (2t + 1)\,dt + C = t^2 + t + C. \] Apply $V(0) = 3$: $3 = 0 + 0 + C$, so $C = 3$, and \[ V(t) = t^2 + t + 3. \] At $t = 4$ minutes: \[ V(4) = 16 + 4 + 3 = 23 \text{ liters}. \]
Check. Differentiate: $V'(t) = 2t + 1$, matches the fill rate. Evaluate at the start: $V(0) = 3$, matches. The answer $23$ liters is above the starting $3$ liters, as predicted, and the units track: the rate is in liters per minute, the volume is in liters.
Common Misconceptions
the rate $g(t)$ in $x' = g(t)$ is itself the answer, so the solution is just $g(t)$.
This confuses the rate of change with the quantity. The function $g(t)$ tells how fast $x$ changes, not what $x$ equals. For $x' = 4t$, the solution is not $4t$; it is $2t^2 + C$, whose derivative is $4t$. The check is always to differentiate the proposed solution and see whether the derivative reproduces $g(t)$. Differentiating $2t^2 + 5$ gives $4t$; differentiating $4t$ gives $4$, which is not $g(t)$.
the initial condition changes the differential equation, so a different starting value gives a different rate law.
The rate law $x' = g(t)$ is fixed. The initial condition does not touch the equation; it only selects which curve from the family is the answer. For $x' = 4t$, the general solution is always $2t^2 + C$. The condition $x(0) = 5$ gives $C = 5$, and a different condition $x(0) = -1$ gives $C = -1$, but the rate law and the family $2t^2 + C$ are the same in both. Solve for the family first, then read the constant off the condition.
omitting $+C$ is harmless because the initial condition will fix it later.
Without $+C$ there is nothing for the initial condition to determine, and the family collapses to a single accidental member. The constant carries the freedom that the initial condition removes. Dropping it on $x' = 6t$ leaves only $3t^2$, which forces $x(0) = 0$ and cannot match $x(0) = 4$. Write $+C$ in every general solution, then solve for it.
an answer left as $\int_{t_0}^{t} g(s)\,ds + x_0$ is unfinished.
When $g$ has no elementary antiderivative, the definite-integral form is the finished, exact solution. The Fundamental Theorem of Calculus (Theorem 1.12) certifies that its derivative is $g(t)$ and that it meets the initial value at $t_0$. Trying to force a closed-form antiderivative for $e^{-t^2}$ wastes effort on a function that provably has none in elementary terms.
Leveled Practice
Level 1: Direct integration
Problem 1. Find the general solution of $x' = 8t^3$.
Show answer
Integrate directly: \[ x(t) = \int 8t^3\,dt + C = 2t^4 + C. \] Differentiating $2t^4 + C$ returns $8t^3$, confirming the answer.
Problem 2. Find the general solution of $x' = \cos t$.
Show answer
\[ x(t) = \int \cos t\,dt + C = \sin t + C. \] The derivative of $\sin t$ is $\cos t$, so the answer checks.
Level 2: Initial value problems with $t_0 = 0$
Problem 3. Solve the IVP $x' = 6t$, $\;x(0) = 2$.
Show answer
General solution: $x(t) = \int 6t\,dt + C = 3t^2 + C$. Apply $x(0) = 2$: \[ 2 = 3(0)^2 + C \implies C = 2. \] Particular solution: $x(t) = 3t^2 + 2$. Check: $x'(t) = 6t$ and $x(0) = 2$, both match.
Problem 4. Solve the IVP $x' = e^{t}$, $\;x(0) = 4$.
Show answer
General solution: $x(t) = \int e^{t}\,dt + C = e^{t} + C$. Apply $x(0) = 4$: \[ 4 = e^{0} + C = 1 + C \implies C = 3. \] Particular solution: $x(t) = e^{t} + 3$. Check: $x'(t) = e^{t}$ and $x(0) = 1 + 3 = 4$, both match.
Level 3: Initial value problems with $t_0 \neq 0$
Problem 5. Solve the IVP $x' = 3t^2$, $\;x(2) = 1$.
Show answer
General solution: $x(t) = \int 3t^2\,dt + C = t^3 + C$. Apply $x(2) = 1$ by substituting $t = 2$, $x = 1$: \[ 1 = (2)^3 + C = 8 + C \implies C = -7. \] Particular solution: $x(t) = t^3 - 7$. Check: $x'(t) = 3t^2$ and $x(2) = 8 - 7 = 1$, both match.
Problem 6. Solve the IVP $x' = \dfrac{1}{t}$, $\;x(1) = 0$, for $t > 0$.
Show answer
General solution for $t > 0$: $x(t) = \int \dfrac{1}{t}\,dt + C = \ln t + C$. Apply $x(1) = 0$: \[ 0 = \ln 1 + C = 0 + C \implies C = 0. \] Particular solution: $x(t) = \ln t$. Check: $x'(t) = \dfrac{1}{t}$ and $x(1) = \ln 1 = 0$, both match. The restriction $t > 0$ keeps the logarithm and the rate $\frac{1}{t}$ defined.
Level 4: Non-elementary integrals and reasoning
Problem 7. Write the solution of the IVP $x' = \dfrac{\sin t}{t}$, $\;x(1) = 2$, as a definite integral, and explain why a closed form is not expected.
Show answer
The integrand $\dfrac{\sin t}{t}$ has no elementary antiderivative, so write the solution with a variable upper limit using the Fundamental Theorem of Calculus: \[ x(t) = \int_{1}^{t} \frac{\sin s}{s}\,ds + 2. \] By Theorem 1.12, $x'(t) = \dfrac{\sin t}{t}$, and at $t = 1$ the integral is zero, so $x(1) = 0 + 2 = 2$. Both conditions hold. This integral defines a special function (the sine integral), which is the reason no elementary formula exists.
Problem 8. Two students solve $x' = 4t$, $\;x(0) = 5$. One integrates first then applies the condition; the other substitutes $t = 0$ into the rate law $x' = 4t$ to get $x'(0) = 0$ and reports $x = 5$. Which student is correct, and what went wrong for the other?
Show answer
The first student is correct: $x(t) = 2t^2 + 5$. The second student confused the rate of change with the quantity. Substituting $t = 0$ into $x' = 4t$ gives the slope at the start, $x'(0) = 0$, not the value of $x$. The value comes from integrating the rate and then using the initial condition, not from evaluating the rate at a single time. Reporting $x = 5$ as a constant function fails the equation, since the derivative of the constant $5$ is $0$, not $4t$.
Level 5: Connecting the two faces of the solution
Problem 9. Show that $\displaystyle x(t) = \int_{t_0}^{t} g(s)\,ds + x_0$ solves the IVP $x' = g(t)$, $\;x(t_0) = x_0$ for any continuous $g$, and explain how this single formula reproduces both the family $\int g(t)\,dt + C$ and the particular solution.
Show answer
The equation holds. By the Fundamental Theorem of Calculus (Theorem 1.12), \[ x'(t) = \frac{d}{dt}\left(\int_{t_0}^{t} g(s)\,ds + x_0\right) = g(t), \] since the derivative of the constant $x_0$ is zero. So $x' = g(t)$.
The initial condition holds. At $t = t_0$ the integral runs from $t_0$ to $t_0$, which is zero, so \[ x(t_0) = \int_{t_0}^{t_0} g(s)\,ds + x_0 = 0 + x_0 = x_0. \]
One formula, both faces. The indefinite-integral family $\int g(t)\,dt + C$ and the definite-integral formula differ only in how the constant is recorded. The definite integral has already absorbed the choice of constant into the lower limit $t_0$ and the value $x_0$: changing the starting point or the starting value shifts the whole curve, which is exactly the role of $C$. One way to see this is that $\int_{t_0}^{t} g(s)\,ds$ is one specific antiderivative of $g$, and adding $x_0$ shifts it to pass through $(t_0, x_0)$. Another way to see it is that the general solution $\int g(t)\,dt + C$ has its constant fixed the moment an initial condition is given, which is what the definite-integral form does from the start. Either path arrives at the same particular solution.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes.
Novice (Levels 1 to 2):
Competent (Levels 3 to 4):
Proficient (Level 5):
Mental Model
A rate law $x' = g(t)$ is a recipe that tells you, at each instant, how fast the quantity is changing, but never where it stands. Recovering the quantity is reading the speedometer backward to find the odometer.
Because the speedometer says nothing about the starting mileage, integrating leaves a free constant: the whole family $\int g(t)\,dt + C$ shares the same rates at every instant but starts at different places. The initial condition is the odometer reading at one known moment. It fixes the constant and selects the single curve that both obeys the recipe and matches the known reading.
When the rate law is too rough for an elementary antiderivative, the same idea still works: the running total $\int_{t_0}^{t} g(s)\,ds$ is the accumulated change from the start time to now, and adding the starting value $x_0$ gives the current amount. The Fundamental Theorem of Calculus is the promise that this running total really does change at the prescribed rate.
Resources
| Resource | What it covers | Link |
|---|---|---|
| Logan, 3rd ed., §1.2 Antiderivatives (pp.16-21) | Theorem 1.12 (FTC), Examples 1.9 through 1.11, the IVP definition | https://www.math.unl.edu/~jlogan1/PDFfiles/New3rdEditionODE.pdf |
| OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations | Initial-value problems and particular solutions, free and openly licensed | https://openstax.org/books/calculus-volume-2/pages/4-1-basics-of-differential-equations |
| Logan 1.2 tutor guide (structure: IVP, FTC, definite versus indefinite integral) | Faculty concept-structure reference, not a graded key | ~/math347-ingest/math_guides/MATH_347/Logan DEs/New Guides/Chapter 1/logan_1_2.tex |
Connections
Within MATH347 (Differential Equations)
- Introduction to differential equations: That lesson defines order, solution, and the initial-value problem in general. Those ideas first become a step-by-step procedure here, on the simplest equation, $x' = g(t)$.
- Separable differential equations: The next solution technique rearranges $x' = f(x)g(t)$ so each variable sits with its own differential, then integrates both sides. The integrate-then-apply-the-condition habit built here is exactly the engine that finishes a separable problem.
Built From
- Antiderivatives and the constant of integration: The general solution of $x' = g(t)$ is an antiderivative, and the $+C$ is the same constant the antiderivative produces.
- Fundamental Theorem of Calculus: The definite-integral form of the solution is correct precisely because differentiating an integral with a variable upper limit returns the integrand.
Audience Notes
For students who find math intimidating: The procedure is short and repeats every time. Integrate the rate, write $+C$, then plug in the one known point to find $C$. A wrong turn here is common and shows something worth understanding; there is no need to rush. If a step feels uncertain, differentiate your answer and see whether it gives back the rate law.
For students interested in proof: Problem 9 is a complete proof that the definite-integral formula solves the IVP for any continuous $g$. It rests entirely on the Fundamental Theorem of Calculus and the fact that an integral over a single point is zero.
For students interested in careers: Recovering a quantity from its rate is the daily work of physics, engineering, and finance. Position from velocity, charge from current, and accumulated cost from a spending rate are all instances of $x' = g(t)$ with a known starting value.
For gifted and curious students: The function $e^{-t^2}$ in Example 4 has no elementary antiderivative, a fact proved by Liouville’s theory of integration in finite terms. Its integral defines the error function, central to probability. The definite-integral form is not a workaround; it is how a large family of named special functions is defined.
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