Initial Value Problems
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.10: “Antiderivatives” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-10-antiderivatives |
| Textbook used in class | Stewart, Calculus, Section 3.9: “Antiderivatives” (Examples 3-4) |
Opening Scenario
Knowing the rate of change of a quantity at every moment is not enough to determine the quantity itself. You also need to know the value at one moment -- the starting point.
If a car is accelerating at $6$ m/s$^2$ at all times but you do not know whether it was initially moving forward or backward, you cannot determine its velocity at any later time. One initial reading pins down the answer completely.
An initial condition is that one reading. Together, the rate-of-change equation and the initial condition form an initial value problem.
Quick Reference
Initial value problem (first order): $$F'(x) = f(x), \quad F(a) = b.$$
Solution method:
- Find the general antiderivative: $F(x) = G(x) + C$ (where $G$ is any one antiderivative of $f$).
- Apply the initial condition: $G(a) + C = b$, so $C = b - G(a)$.
- Write the particular solution: $F(x) = G(x) + (b - G(a))$.
Key Concepts
1. First-Order Initial Value Problem
Example 1. Find the function $f$ such that $f'(x) = 3x^2 + 1$ and $f(0) = 5$.
Step 1: General antiderivative. $f(x) = x^3 + x + C$.
Step 2: Apply the initial condition. $f(0) = 0 + 0 + C = 5 \Rightarrow C = 5$.
Step 3: Particular solution. $f(x) = x^3 + x + 5$.
Verify: $f'(x) = 3x^2 + 1$ and $f(0) = 0 + 0 + 5 = 5$. Confirmed.
Boxed answer: $f(x) = x^3 + x + 5$.
2. Second-Order Initial Value Problem
Sometimes you are given the second derivative $f''$ and two initial conditions: $f(a)$ and $f'(a)$. Antidifferentiate twice, finding $C_1$ and $C_2$ from the two conditions.
Example 2. Find $f$ given $f''(x) = 6x + 4$, $f(0) = 2$, and $f'(0) = 1$.
Step 1: First antiderivative ($f''$ to $f'$). $f'(x) = 3x^2 + 4x + C_1$.
Apply $f'(0) = 1$: $0 + 0 + C_1 = 1 \Rightarrow C_1 = 1$.
So $f'(x) = 3x^2 + 4x + 1$.
Step 2: Second antiderivative ($f'$ to $f$). $f(x) = x^3 + 2x^2 + x + C_2$.
Apply $f(0) = 2$: $0 + 0 + 0 + C_2 = 2 \Rightarrow C_2 = 2$.
Particular solution: $f(x) = x^3 + 2x^2 + x + 2$.
Boxed answer: $f(x) = x^3 + 2x^2 + x + 2$.
Recap. Each integration introduces one constant; each initial condition pins down one constant. A second-order IVP (with $f''$ given and two conditions) requires two integrations and two conditions.
3. Finding C with a Non-Zero Starting Point
Example 3. Find $g$ given $g'(x) = \cos x$ and $g(\pi) = 3$.
General antiderivative: $g(x) = \sin x + C$.
Apply the condition: $g(\pi) = \sin(\pi) + C = 0 + C = 3 \Rightarrow C = 3$.
Particular solution: $g(x) = \sin x + 3$.
Boxed answer: $g(x) = \sin x + 3$.
“$g(\pi) = \sin(\pi) + C$ gives $C = 3 - \sin(\pi)$, but since $\sin(\pi) = 0$ anyway, I should just write $C = 3$ without checking.” Actually, always substitute and simplify step by step. For the $\sin$ function, $\sin(\pi) = 0$ is simple, but for other functions the evaluation is less obvious. Practice the full substitution every time.
4. When the Initial Condition Is Not at Zero
Example 4. Find $h$ such that $h'(x) = e^x$ and $h(2) = 1$.
$h(x) = e^x + C$.
$h(2) = e^2 + C = 1 \Rightarrow C = 1 - e^2$.
$h(x) = e^x + 1 - e^2$.
Boxed answer: $h(x) = e^x + (1 - e^2)$.
Note that the constant $1 - e^2 \approx 1 - 7.39 \approx -6.39$ is a specific negative number, not something to simplify further.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting $C$ entirely in the general antiderivative | Writing $F(x) = x^3$ before applying the condition | Write $F(x) = x^3 + C$ first; the condition determines $C$ |
| Applying the condition to $f'$ instead of $f$ | For $f'(x) = 3x^2$, applying $f(0) = 2$ as $f'(0) = 2$ | The condition gives the value of $f$, not $f'$, at the given point |
| Using two constants but only one condition | Writing $f(x) = x^3 + C_1 + C_2$ with one condition | One integration introduces one constant; a second integration introduces a second |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find $f$ such that $f'(x) = x^2 - x$ and $f(1) = 0$.
Show answer
General antiderivative: $f(x) = \dfrac{x^3}{3} - \dfrac{x^2}{2} + C$.
$f(1) = \dfrac{1}{3} - \dfrac{1}{2} + C = 0 \Rightarrow C = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$.
Boxed answer: $f(x) = \dfrac{x^3}{3} - \dfrac{x^2}{2} + \dfrac{1}{6}$.
Level 2 -- Multiple Steps
Problem 2. Find $f$ such that $f''(x) = \sin x$, $f'(0) = 2$, and $f(0) = -1$.
Show answer
First integration: $f'(x) = -\cos x + C_1$. $f'(0) = -1 + C_1 = 2 \Rightarrow C_1 = 3$.
$f'(x) = -\cos x + 3$.
Second integration: $f(x) = -\sin x + 3x + C_2$. $f(0) = 0 + 0 + C_2 = -1 \Rightarrow C_2 = -1$.
Boxed answer: $f(x) = -\sin x + 3x - 1$.
Level 3 -- Deeper Problems
Problem 3. Is there more than one function satisfying $f'(x) = 2x$ and $f(1) = 3$? Justify.
Show answer
The general antiderivative of $2x$ is $x^2 + C$. The initial condition $f(1) = 1 + C = 3$ gives $C = 2$. So the particular solution is $f(x) = x^2 + 2$.
Is it unique? By the theorem that all antiderivatives of $f$ differ by a constant, any antiderivative of $2x$ is $x^2 + C$ for some $C$. The condition $f(1) = 3$ forces $C = 2$. Since $C$ is uniquely determined, the particular solution is unique.
Boxed answer: No. There is exactly one function satisfying both conditions: $f(x) = x^2 + 2$.
Mastery Checklist
Mental Model
An initial value problem is a trajectory-pinning problem. The differential equation $F' = f$ specifies the shape of the family of trajectories (all parallel curves with the same slope at each $x$). The initial condition selects the unique trajectory that passes through the given point $(a, b)$.
One condition pins one constant. Two conditions pin two constants. The number of conditions needed equals the number of times you integrate.
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