4.2 Solving Simple Initial-Value Problems in Mechanics

Infinitely Many Futures

Drop a ball from your hand. It falls, accelerating at $g = 9.8 \text{ m/s}^2$. The differential equation governing its velocity is

$$\frac{dv}{dt} = -g$$

This equation says one thing: at every instant, the velocity is decreasing at a rate of $9.8 \text{ m/s}^2$. It is a rule about how velocity changes --- a motion-generating recipe, as we discussed in Section 4.1.

But here is something unsettling. The equation $\frac{dv}{dt} = -g$ has infinitely many solutions. An object dropped from rest satisfies it. An object thrown upward at $20 \text{ m/s}$ satisfies it. An object already falling at $50 \text{ m/s}$ satisfies it. Every one of these motions obeys the same acceleration law --- the same differential equation --- yet they produce completely different trajectories.

The ball you are holding in your hand right now, the one you are about to drop, will follow exactly one of those infinitely many solutions. What extra information picks out the right one?

Before you read on: Two students solve the same differential equation and both get $v(t) = Ce^{-t}$. One says $C = 5$, the other says $C = -3$. Can they both be correct?

(a) No --- there should be one unique solution to any differential equation

(b) Yes --- they might be solving for different physical situations

(c) Only if one of them made a mistake

Commit to your answer before continuing.

[Interactive: Predict-Then-Reveal. Student selects one of three choices. After submitting, the correct answer is revealed: (b). A brief explanation appears: "Both students are correct. The general solution $v(t) = Ce^{-t}$ represents a whole family of solutions --- one for each value of $C$. The student with $C = 5$ is describing an object that starts with velocity $5$ at $t = 0$. The student with $C = -3$ is describing an object that starts with velocity $-3$. Same equation, different starting conditions, different specific solutions."]

If you chose (b), good --- you are already sensing the key idea. A differential equation does not describe one motion. It describes a family of motions. Something else is needed to select a single member of that family.

Quick Recall: Section 4.1

Before we move forward, let's make sure the foundations from the previous section are solid.

Recall prompt: In Section 4.1, we learned that a statement like "drag is proportional to speed" translates into a differential equation. Can you write that DE? And what did we mean by saying a DE is a "motion-generating rule"?

If those feel uncertain, revisit Section 4.1 before continuing. Everything here builds on the idea that a differential equation encodes how a system evolves --- but not where it starts.

Exploration: A Family of Curves

The best way to understand why initial conditions matter is to see a whole family of solutions at once.

[Interactive: Solution Family Explorer. The screen shows a $(t, x)$ coordinate plane with axes from $t = 0$ to $t = 5$ and $x = -3$ to $x = 3$. Dozens of solution curves for the differential equation $\frac{dx}{dt} = -x$ are drawn in light gray --- exponential decays starting from various positive and negative values, all approaching zero. When the student clicks anywhere in the $(t, x)$ plane, the specific solution curve passing through that point lights up in bold blue, and its equation appears: $x(t) = x_0 \, e^{-t}$, with the value of $x_0$ filled in. A readout below shows: "You selected the initial condition $x(0) = __$. The specific solution is $x(t) = __ \, e^{-t}$."

Guided prompts appear as the student explores:]

Prompt 1: Click on the point $(0, 2)$. What solution lights up? Now click on $(0, -1)$. How does the curve change? What stays the same about its shape?

Prompt 2: Click on several different points, all at $t = 0$ but at different heights. Each click selects a different solution. What role does the $x$-coordinate of your click play?

Prompt 3: Now click on a point that is not at $t = 0$ --- say, the point $(1, 1)$. Does the system still find a unique curve through that point? What initial condition does it correspond to?

Prompt 4: Try to find two different solution curves that cross each other. Can you? Why or why not?

If you spent time with that interactive, you discovered several things:

Every point in the $(t, x)$ plane lies on exactly one solution curve. No two solution curves cross.
Clicking at $t = 0$ directly selects the initial condition. Clicking at a later time still determines a unique curve, but the initial condition must be calculated backward.
All the curves have the same shape --- they are all exponential decays toward zero. The initial condition controls which particular decay you are on, but not the character of the decay itself.

This is the central geometric picture: the differential equation fills the plane with a family of curves. The initial condition selects one.

Concept Reveal: The Initial-Value Problem

Here is the idea, stated plainly.

A differential equation like $\frac{dv}{dt} = -g$ defines a family of possible motions --- all the ways an object could move while obeying that acceleration law. This family contains infinitely many members, one for every possible starting state.

An initial condition specifies the state of the system at a particular moment --- typically at $t = 0$. For a velocity equation, this means specifying $v(0) = v_0$. For a position equation, it means specifying $x(0) = x_0$.

Together, the differential equation and the initial condition form an initial-value problem (IVP):

$$\frac{dv}{dt} = -g, \qquad v(0) = v_0$$

The IVP is the complete mathematical model of a specific motion. The DE provides the rule; the initial condition provides the starting point. Together, they determine a unique solution --- one curve out of the family.

This is worth saying plainly: a differential equation alone does not predict motion. An initial-value problem does. The DE tells you the law. The initial condition tells you which specific situation you are in. You need both.

Pause and connect: In Chapter 2, every kinematic problem gave you $x_0$ and $v_0$ --- a starting position and a starting velocity. Those were initial conditions. You used them every time you plugged into $x(t) = x_0 + v_0 t + \frac{1}{2}at^2$. The concept is not new. What is new is the name: you were solving initial-value problems all along. Now we are naming the structure you have already been using.

Worked Example: Constant Acceleration ($dv/dt = -g$)

Let's solve our first IVP with all steps visible. This example will feel familiar from Chapter 2, and that is the point --- we are re-deriving a result you already know, using the new language of differential equations.

Problem: An object is dropped from rest. Find its velocity as a function of time.

The IVP:

$$\frac{dv}{dt} = -g, \qquad v(0) = 0$$

Step 1: Integrate both sides with respect to $t$.

The equation $\frac{dv}{dt} = -g$ says the derivative of $v$ is a constant. To recover $v$, we integrate:

$$\int \frac{dv}{dt} \, dt = \int (-g) \, dt$$

The left side is simply $v(t)$ (by the fundamental theorem of calculus). The right side is $-gt + C$, where $C$ is an arbitrary constant of integration:

$$v(t) = -gt + C$$

Step 2: Apply the initial condition.

We know $v(0) = 0$. Substituting $t = 0$:

$$v(0) = -g(0) + C = C$$

So $C = 0$.

Step 3: Write the specific solution.

$$v(t) = -gt$$

Step 4: Interpret.

The constant $C$ was not just a mathematical artifact. It was the initial velocity. When $C = 0$, the object starts from rest. If we had set $v(0) = v_0$ instead, we would have found $C = v_0$, giving $v(t) = v_0 - gt$ --- exactly the formula from Section 2.2.

The constant of integration is a placeholder for physical information that the DE alone does not contain. The initial condition fills it in.

Pause and think: In this example, the "family of solutions" is the set of all lines with slope $-g$: $v(t) = -gt + C$ for every value of $C$. Each line represents an object falling under gravity, but starting at a different velocity. The initial condition selects one line from this family.

Finding Position: A Two-Step IVP

Once you have $v(t)$, you can find position by solving a second IVP.

Problem: The same object is dropped from rest at height $x_0 = 20 \text{ m}$ (taking upward as positive). Find $x(t)$.

The IVP:

$$\frac{dx}{dt} = v(t) = -gt, \qquad x(0) = 20$$

Step 1: Integrate.

$$x(t) = \int (-gt) \, dt = -\frac{1}{2}gt^2 + C$$

Step 2: Apply the initial condition.

$$x(0) = -\frac{1}{2}g(0)^2 + C = C = 20$$

Step 3: Write the specific solution.

$$x(t) = 20 - \frac{1}{2}gt^2$$

This is exactly $x(t) = x_0 + v_0 t + \frac{1}{2}at^2$ with $v_0 = 0$, $a = -g$, and $x_0 = 20$. The kinematic equation from Chapter 2 was the solution to this IVP all along.

Notice the pattern: each integration introduces one constant, and each initial condition determines one constant. A second-order problem (going from acceleration to position) requires two integrations and two initial conditions --- $v_0$ and $x_0$ --- to produce a unique solution. This is why every problem in Chapter 2 gave you both a starting velocity and a starting position.

Faded Example: Velocity-Dependent Acceleration ($dv/dt = -bv$)

Now let's try something that you could not do with the constant-acceleration formulas. This is the equation for a velocity-dependent drag force (Section 4.1): an object whose acceleration is proportional to its velocity.

Problem: Solve the IVP

$$\frac{dv}{dt} = -bv, \qquad v(0) = v_0$$

where $b > 0$ is a constant.

This equation is different from $dv/dt = -g$ in an important way: the right side depends on $v$, the unknown function. You cannot simply integrate both sides with respect to $t$ because $v$ appears on the right. Instead, we use a technique called separation of variables.

Step 1: Separate the variables.

Rewrite the equation so that everything involving $v$ is on one side and everything involving $t$ is on the other. Start by dividing both sides by $v$ (assuming $v \neq 0$):

$$\frac{1}{v}\frac{dv}{dt} = -b$$

Now "multiply both sides by $dt$" (treating $\frac{dv}{dt}$ as a ratio of differentials --- this is a notational shorthand that can be made rigorous, but for now treat it as a reliable computational tool):

$$\frac{dv}{v} = -b \, dt$$

The variables are now separated: $v$ on the left, $t$ on the right.

Step 2: Integrate both sides.

$$\int \frac{dv}{v} = \int (-b) \, dt$$

$$\ln |v| = -bt + C_1$$

where $C_1$ is a constant of integration. (We will clean up the absolute value in a moment.)

Step 3: Solve for $v$.

Exponentiate both sides:

$$|v| = e^{-bt + C_1} = e^{C_1} \cdot e^{-bt}$$

Since $e^{C_1}$ is a positive constant, let us call it $A$:

$$v(t) = \pm A \, e^{-bt}$$

The $\pm$ and the constant $A$ can be combined into a single constant that we will call $C$ (which can be positive or negative):

$$v(t) = C \, e^{-bt}$$

This is the general solution --- a family of exponential decays, one for each value of $C$.

Step 4: Apply the initial condition. (Your turn.)

Before you read on: You have $v(t) = Ce^{-bt}$ and the initial condition $v(0) = v_0$. Determine the value of $C$.

[Interactive: Predict-Then-Reveal. A text input field where the student types their answer for $C$. After submitting, the solution is shown.]

Setting $t = 0$:

$$v(0) = Ce^{-b \cdot 0} = C \cdot 1 = C$$

So $C = v_0$.

Step 5: Write the specific solution.

$$\boxed{v(t) = v_0 \, e^{-bt}}$$

Step 6: Interpret.

The velocity decays exponentially from its initial value $v_0$ toward zero. The object slows down continuously but never fully stops. After a time $t = 1/b$, the velocity has dropped to $v_0 / e \approx 0.37\, v_0$ --- about 37% of its initial value. The constant $b$ controls how quickly the decay happens: large $b$ means rapid deceleration.

The constant of integration $C$ turned out to be the initial velocity --- just as it was in the constant-acceleration case. This is not a coincidence. The constant of integration always carries the information that the initial condition provides. It is the mathematical slot where the specific physical situation gets plugged in.

Independent Problem: Falling with Linear Drag ($dv/dt = g - bv$)

Now it is your turn. This problem combines gravity and drag --- an object falling through a resistive medium.

Problem: Solve the IVP

$$\frac{dv}{dt} = g - bv, \qquad v(0) = 0$$

where $g$ and $b$ are positive constants, and downward is positive.

This models an object dropped from rest in a medium that exerts a drag force proportional to velocity. At $t = 0$, the object is not moving, so there is no drag and the acceleration is just $g$. As the object speeds up, the drag term $bv$ grows, reducing the net acceleration.

Try it yourself before looking at the solution. Here is a roadmap:

Separate the variables: get $v$ on one side, $t$ on the other.
Integrate both sides. (Hint: $\int \frac{dv}{g - bv}$ requires a substitution or recognition of the form $\int \frac{du}{a - bu}$.)
Solve for $v(t)$.
Apply the initial condition $v(0) = 0$ to find the constant.

Check your answer

**Step 1: Separate variables.** $$\frac{dv}{g - bv} = dt$$ **Step 2: Integrate both sides.** The left side requires recognizing the integral $\int \frac{dv}{g - bv}$. Let $u = g - bv$, so $du = -b \, dv$, which gives $dv = -\frac{du}{b}$: $$\int \frac{dv}{g - bv} = \int \frac{-du/b}{u} = -\frac{1}{b} \ln |u| = -\frac{1}{b} \ln |g - bv|$$ So: $$-\frac{1}{b} \ln |g - bv| = t + C_1$$ **Step 3: Solve for $v$.** Multiply both sides by $-b$: $$\ln |g - bv| = -bt - bC_1$$ Exponentiate: $$|g - bv| = e^{-bC_1} \cdot e^{-bt}$$ Let $A = e^{-bC_1}$ (a positive constant), and absorb the sign: $$g - bv = A\, e^{-bt}$$ Solve for $v$: $$v(t) = \frac{g}{b} - \frac{A}{b}\, e^{-bt}$$ **Step 4: Apply the initial condition.** $v(0) = 0$: $$0 = \frac{g}{b} - \frac{A}{b} \cdot 1$$ So $A = g$, and the specific solution is: $$\boxed{v(t) = \frac{g}{b}\left(1 - e^{-bt}\right)}$$ **Interpretation:** At $t = 0$, the exponential term is $1$, so $v = 0$ --- the object starts from rest. As $t \to \infty$, the exponential term vanishes, and $v \to g/b$. This limiting velocity is called the **terminal velocity**: the speed at which gravitational acceleration and drag exactly balance, producing zero net acceleration. The quantity $g/b$ has units of velocity (m/s), and $1/b$ has units of time (s). The time $1/b$ is roughly how long it takes to approach terminal velocity --- after $t = 3/b$, the velocity is already within 5% of terminal speed. Notice the structure: the solution is a constant (the terminal velocity) minus a decaying exponential. The initial condition set the coefficient of the exponential to $g$, ensuring the solution starts at zero.

The Physical Meaning of Constants of Integration

Let's pause and reflect on what we have seen across these three examples, because there is a pattern worth naming explicitly.

IVP	General solution	What $C$ turned out to be
$\frac{dv}{dt} = -g$	$v = -gt + C$	$C = v_0$ (initial velocity)
$\frac{dv}{dt} = -bv$	$v = Ce^{-bt}$	$C = v_0$ (initial velocity)
$\frac{dv}{dt} = g - bv$	$v = \frac{g}{b} + Ce^{-bt}$	$C = -\frac{g}{b}$ (set by $v_0 = 0$)

In every case, the constant of integration was not an abstract mathematical artifact. It was a stand-in for measurable physical information --- information that the differential equation does not supply but that the initial condition does.

This is worth saying plainly: every constant of integration in a physics problem has a physical meaning. It corresponds to something you could measure about the system at a specific moment. When you "apply the initial condition," you are not performing a mathematical trick. You are plugging in a measurement.

The DE says: here is the rule that governs this system. The initial condition says: here is the state the system was actually in when we started watching. The two together produce a complete, unique prediction.

Multiple Representations: Seeing the IVP Geometrically

We have been working algebraically. Let's see the same ideas graphically and numerically.

The graphical view

Consider the family of solutions to $\frac{dv}{dt} = -bv$: the curves $v(t) = Ce^{-bt}$ for various values of $C$.

[Interactive: IVP Graphical View. A $(t, v)$ plane shows multiple solution curves for $v(t) = Ce^{-bt}$ with $b = 1$: curves for $C = -3, -2, -1, 0, 1, 2, 3$ are drawn in light gray. A slider labeled "$v_0$" runs from $-3$ to $3$. As the student moves the slider, one curve highlights in bold, and the label updates: "Initial condition: $v(0) = v_0$. Solution: $v(t) = v_0 \, e^{-t}$." A second slider labeled "$b$" (from $0.2$ to $3$) changes the decay rate, causing all curves to steepen or flatten simultaneously.]

Prompt 1: Set $b = 1$ and sweep $v_0$ from $-3$ to $3$. What do all these curves have in common? What is different?

Prompt 2: Now fix $v_0 = 2$ and vary $b$. How does the decay rate change? At what time has the velocity dropped to half its initial value?

Prompt 3: Is there any solution that increases over time? Why or why not?

The graphical view makes something vivid that the algebra can obscure: the initial condition slides you along the family of curves, selecting one. The parameter $b$ in the differential equation changes the shape of all the curves simultaneously. The initial condition selects which curve you are on; the DE determines what all the curves look like.

The numerical view

Here is a simple table for $v(t) = 5e^{-t}$ (so $v_0 = 5$, $b = 1$):

$t$ (s)	$v(t)$ (m/s)	$dv/dt$ (m/s$^2$)	Is $dv/dt = -v$?
0	5.00	$-5.00$	$\checkmark$
1	1.84	$-1.84$	$\checkmark$
2	0.68	$-0.68$	$\checkmark$
3	0.25	$-0.25$	$\checkmark$
4	0.09	$-0.09$	$\checkmark$

At every row, the rate of change $dv/dt$ equals $-v$. The solution satisfies the differential equation at every instant. And the first row confirms the initial condition: $v(0) = 5$.

Pause and think: If a different student had constructed a table for $v(t) = -3e^{-t}$ (initial condition $v_0 = -3$), would the last column still show check marks? Try a few rows mentally.

Yes --- because $-3e^{-t}$ also satisfies $dv/dt = -v$. The DE is the same; only the initial condition (and therefore the specific numbers) differ.

Practice Layers

Layer 1: Concrete --- Solve IVPs for constant and linear acceleration models

Problem 1. Solve the IVP:

$$\frac{dv}{dt} = 5, \qquad v(0) = 3$$

Check your answer

Integrate: $v(t) = 5t + C$. Apply the initial condition: $v(0) = C = 3$. Solution: $v(t) = 5t + 3$. The velocity increases linearly from $3 \text{ m/s}$ at a rate of $5 \text{ m/s}^2$. This is uniform acceleration --- the same type of problem you solved in Chapter 2, but now framed as an IVP.

Problem 2. Solve the IVP:

$$\frac{dx}{dt} = 4 - 2t, \qquad x(0) = 10$$

Check your answer

Integrate: $x(t) = 4t - t^2 + C$. Apply the initial condition: $x(0) = C = 10$. Solution: $x(t) = 10 + 4t - t^2$. The object starts at $x = 10$ m and initially moves in the positive direction (since $dx/dt = 4$ at $t = 0$). The velocity decreases linearly, reaching zero at $t = 2$ s, after which the object reverses direction. The position reaches its maximum value of $x(2) = 10 + 8 - 4 = 14$ m at $t = 2$ s.

Problem 3. Solve the IVP using separation of variables:

$$\frac{dv}{dt} = -2v, \qquad v(0) = 8$$

Check your answer

**Separate:** $\frac{dv}{v} = -2 \, dt$. **Integrate:** $\ln|v| = -2t + C_1$. **Solve for $v$:** $v = Ce^{-2t}$. **Apply initial condition:** $v(0) = C = 8$. **Solution:** $v(t) = 8e^{-2t}$. The velocity decays exponentially from $8 \text{ m/s}$ with a time constant of $1/2 = 0.5$ s. After $0.5$ s, the velocity is $8e^{-1} \approx 2.94 \text{ m/s}$ --- about 37% of its initial value. After $1$ s, it is $8e^{-2} \approx 1.08 \text{ m/s}$. The object slows rapidly but never fully stops.

Layer 2: Pattern --- Determine constants from initial conditions

In each problem below, you are given a general solution and an initial condition. Find the specific solution.

Problem 4. General solution: $x(t) = A\cos(3t) + B\sin(3t)$. Initial conditions: $x(0) = 4$, $\frac{dx}{dt}(0) = 0$.

Check your answer

Apply $x(0) = 4$: $$x(0) = A\cos(0) + B\sin(0) = A$$ So $A = 4$. Now find $\frac{dx}{dt}$: $$\frac{dx}{dt} = -3A\sin(3t) + 3B\cos(3t)$$ Apply $\frac{dx}{dt}(0) = 0$: $$\frac{dx}{dt}(0) = -3A\sin(0) + 3B\cos(0) = 3B$$ So $3B = 0$, meaning $B = 0$. **Specific solution:** $x(t) = 4\cos(3t)$. This describes an oscillation with amplitude 4, starting at the maximum displacement with zero initial velocity --- like pulling a spring to $x = 4$ and releasing it from rest.

Problem 5. General solution: $v(t) = \frac{g}{b} + Ce^{-bt}$. Initial condition: $v(0) = 10 \text{ m/s}$ with $g = 9.8 \text{ m/s}^2$ and $b = 0.5 \text{ s}^{-1}$.

Check your answer

The terminal velocity is $g/b = 9.8 / 0.5 = 19.6 \text{ m/s}$. Apply $v(0) = 10$: $$10 = 19.6 + C$$ So $C = 10 - 19.6 = -9.6$. **Specific solution:** $v(t) = 19.6 - 9.6\,e^{-0.5t}$ m/s. This is an object already moving at $10 \text{ m/s}$ that accelerates toward its terminal velocity of $19.6 \text{ m/s}$. It starts below terminal speed, so it speeds up. The exponential term decays, and the velocity asymptotically approaches $19.6 \text{ m/s}$. Compare this to the dropped-from-rest case, where $C = -19.6$. Same DE, same terminal velocity, but a different initial condition changes the transient behavior.

Problem 6. General solution: $x(t) = C_1 e^{2t} + C_2 e^{-2t}$. Initial conditions: $x(0) = 0$, $\frac{dx}{dt}(0) = 4$.

Check your answer

Apply $x(0) = 0$: $$0 = C_1 + C_2$$ So $C_2 = -C_1$. Find $\frac{dx}{dt}$: $$\frac{dx}{dt} = 2C_1 e^{2t} - 2C_2 e^{-2t}$$ Apply $\frac{dx}{dt}(0) = 4$: $$4 = 2C_1 - 2C_2 = 2C_1 - 2(-C_1) = 4C_1$$ So $C_1 = 1$ and $C_2 = -1$. **Specific solution:** $x(t) = e^{2t} - e^{-2t}$. This can also be written as $x(t) = 2\sinh(2t)$, using the hyperbolic sine function. The object starts at the origin with velocity $4$ and accelerates away from it. Without the initial conditions, you would not know whether the motion was symmetric ($C_1 = C_2$), started from rest ($C_1 = -C_2$), or something else entirely. Two constants of integration required two initial conditions to pin down.

Layer 3: Structure --- What changes, what stays the same?

These questions ask you to think about the roles of the DE and the initial condition.

Problem 7. Consider the IVP $\frac{dv}{dt} = -v$, $v(0) = v_0$, with solution $v(t) = v_0 e^{-t}$.

(a) What happens to the solution if you change only the initial condition (say, from $v_0 = 5$ to $v_0 = 10$)?

(b) What happens if you change only the DE (say, from $\frac{dv}{dt} = -v$ to $\frac{dv}{dt} = -3v$), keeping $v_0 = 5$?

Check your answer

(a) **Changing only the initial condition:** The solution changes from $v = 5e^{-t}$ to $v = 10e^{-t}$. The shape of the curve is the same --- still an exponential decay with time constant $1$ s. But the curve is stretched vertically. The initial condition controls the *amplitude* of the solution, not its character. (b) **Changing only the DE:** The solution changes from $v = 5e^{-t}$ to $v = 5e^{-3t}$. The starting value is the same ($v_0 = 5$), but the decay is three times faster. After $1$ second, the first solution gives $v = 5e^{-1} \approx 1.84$, while the second gives $v = 5e^{-3} \approx 0.25$. The DE controls the *shape* and *character* of the solution --- the rate at which it decays. (c) **Changing both:** The solution becomes $v = 10e^{-3t}$. Both the amplitude and the decay rate change. The DE and the initial condition play different, independent roles: the DE determines the *type* of motion; the initial condition selects the *specific instance* of that type.

Problem 8. Two objects fall through the same medium (same $b$) but are released with different initial velocities: $v_0 = 0$ and $v_0 = 2g/b$ (twice the terminal velocity). The solution is $v(t) = \frac{g}{b} + \left(v_0 - \frac{g}{b}\right)e^{-bt}$.

Describe the motion of each object qualitatively. Do they eventually move the same way?

Check your answer

For $v_0 = 0$: The solution is $v(t) = \frac{g}{b}(1 - e^{-bt})$. The object starts from rest and accelerates toward the terminal velocity $g/b$, approaching it from below. The acceleration is initially $g$ and decreases to zero. For $v_0 = 2g/b$: The solution is $v(t) = \frac{g}{b} + \frac{g}{b}e^{-bt} = \frac{g}{b}(1 + e^{-bt})$. The object starts at twice the terminal velocity and *decelerates* toward $g/b$, approaching it from above. The drag initially exceeds gravity, producing a negative net acceleration (the object slows down). Both objects approach the same terminal velocity $g/b$ as $t \to \infty$. The exponential transient dies away regardless of its sign or magnitude. The initial condition determines the *transient* behavior (how the system starts), while the DE determines the *long-term* behavior (where the system ends up). This is a profound structural insight: **for this type of DE, the initial condition is eventually forgotten.** The system's long-term fate is determined by the equation, not by how it started. Not all DEs behave this way --- but this one does, and understanding why is the beginning of a much deeper story.

Layer 4: Transfer --- Same structure, different domain

Problem 9. A bank account earns interest at a rate proportional to its current balance:

$$\frac{dB}{dt} = rB$$

where $B(t)$ is the balance at time $t$ (in years) and $r$ is the annual interest rate. Solve with the initial condition $B(0) = B_0$ (the initial deposit).

Check your answer

This has exactly the same structure as $\frac{dv}{dt} = -bv$, except with a positive constant instead of a negative one. Separation of variables gives: **Separate:** $\frac{dB}{B} = r \, dt$ **Integrate:** $\ln|B| = rt + C_1$ **Solve for $B$:** $B(t) = Ce^{rt}$ **Apply initial condition:** $B(0) = C = B_0$ **Solution:** $B(t) = B_0 \, e^{rt}$ The balance grows exponentially from the initial deposit. After time $t = \ln(2)/r$, the balance has doubled. The interest rate $r$ plays the same role as the decay constant $b$ in the drag problem --- it controls the *rate* of exponential change. The only difference is the sign: $b > 0$ gives decay, $r > 0$ gives growth. The mathematical structure is identical. A biologist studying bacterial growth, an economist modeling inflation, and a physicist studying radioactive decay all solve the same IVP. The letters change. The mathematics does not.

Problem 10. A cup of hot coffee cools according to Newton's law of cooling:

$$\frac{dT}{dt} = -k(T - T_{\text{room}})$$

where $T$ is the coffee's temperature, $T_{\text{room}} = 20°\text{C}$ is the room temperature, and $k > 0$ is a constant. At $t = 0$, the coffee is at $T_0 = 90°\text{C}$. Find $T(t)$.

Check your answer

Let $u = T - T_{\text{room}}$, so $\frac{du}{dt} = \frac{dT}{dt} = -ku$. This is the familiar exponential decay equation. **Solve:** $u(t) = Ce^{-kt}$, so $T(t) = T_{\text{room}} + Ce^{-kt}$. **Apply initial condition:** $T(0) = T_{\text{room}} + C = 90$, so $C = 90 - 20 = 70$. **Solution:** $T(t) = 20 + 70\,e^{-kt}$ (in degrees Celsius). The coffee temperature decays exponentially from $90°\text{C}$ toward the room temperature $20°\text{C}$, never quite reaching it. This has the same structure as the falling-with-drag problem: the solution approaches a steady state ($T_{\text{room}}$ instead of $g/b$) via an exponential transient whose initial amplitude ($70°\text{C}$ instead of $g/b$) is set by the initial condition. The DE determines the *equilibrium* (room temperature) and the *decay rate* ($k$). The initial condition determines *how far from equilibrium you start*.

Reflection

Think about the three examples we worked through in detail: constant acceleration, velocity-proportional drag, and gravity with drag. In every case, the differential equation alone produced infinitely many possible motions. Only the initial condition narrowed it to one.

Why do physicists care about initial conditions so much?

Here is one way to think about it. The laws of physics --- encoded in the differential equations --- are universal. Gravity pulls on everything the same way. Drag acts on every object moving through a fluid. But the universe is full of objects in different states: different positions, different velocities, different temperatures. The laws are the same everywhere; what makes each situation unique is the initial condition.

The equations of motion tell you how the universe evolves. The initial conditions tell you which universe you are in.

This is the essence of determinism in classical mechanics: given the law and the initial state, the entire future (and past) is determined. Nothing is left to chance. Newton's great insight was not just the laws themselves --- it was the realization that laws plus initial conditions are enough.

Looking Ahead

You now know how to solve an initial-value problem: integrate the DE to get a general solution, then use the initial condition to pin down the constants. The specific form of the DE --- whether the right side is a constant, a function of $t$, or a function of $v$ --- determines the technique and the character of the solution.

In Section 4.3, we will systematically compare the three most important acceleration model families: constant acceleration ($a = \text{const}$), velocity-proportional acceleration ($a = -bv$), and gravity-plus-drag ($a = g - bv$). You have already solved the IVPs for all three. Now we will place them side by side and ask: how does the form of the acceleration law shape the character of the motion? You will see that constant acceleration produces parabolas, velocity-proportional drag produces exponential decay, and gravity-with-drag produces approach to terminal velocity --- and that you can predict this behavior from the structure of the equation before you ever solve it.

The mathematical machinery you built in this section --- separation of variables, applying initial conditions, interpreting constants of integration --- will be the engine that drives that comparison.