8.4 Separable Differential Equations & Growth/Decay Models

8.4 Separable Differential Equations & Growth/Decay Models

Core CLEP Concept: A differential equation is an equation relating an unknown function to one or more of its derivatives. On the CLEP Calculus exam, differential equation questions focus on first-order separable equations, initial value problems (IVPs), the universal exponential growth and decay model ($\frac{dy}{dt} = ky$), Newton's Law of Cooling, and graphical interpretation of slope fields. Mastering separation of variables and logarithmic/exponential manipulations ensures maximum points on both calculator and non-calculator sections.

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1. Foundations: Terminology & Solution Types

Definitions & Terminology

• Differential Equation (DE): An equation containing derivatives of an unknown function (e.g., $\frac{dy}{dx} = 2xy$).
• Order of a DE: The order of the highest derivative present. On the CLEP exam, all tested differential equations are first-order (involving only first derivatives $\frac{dy}{dx}$ or $\frac{dy}{dt}$).
• General Solution: A family of functions containing an arbitrary constant of integration $C$ that satisfies the differential equation across its domain.
• Initial Condition: A specified coordinate pair $(x_0, y_0)$ or $(t_0, y_0)$ defining the state of the system at a known instant.
• Particular Solution: The single, unique curve from the general solution family that passes through the given initial condition point.

Verifying Candidate Solutions

To verify whether a candidate function $y = f(x)$ solves a given differential equation:

1. Compute the derivative $\frac{dy}{dx} = f'(x)$.
2. Substitute $y$ and $\frac{dy}{dx}$ into both sides of the differential equation.
3. Confirm that the equation simplifies to a mathematical identity ($0 = 0$ or LHS $\equiv$ RHS).

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2. The Method of Separation of Variables

A first-order differential equation is separable if it can be written in the form:

$\frac{dy}{dx} = g(x) \cdot h(y) \quad \text{or} \quad \frac{dy}{dx} = \frac{g(x)}{p(y)}$

The 4-Step Separation Protocol

+-------------------------------------------------------------------------+
|                 SEPARATION OF VARIABLES 4-STEP PROTOCOL                 |
+-------------------------------------------------------------------------+
| 1. SEPARATE: Move all y terms and dy to LHS; all x terms and dx to RHS. |
|              (1 / h(y)) dy = g(x) dx   [Differentials MUST be in tops!] |
| 2. INTEGRATE: Integrate both sides: int (1 / h(y)) dy = int g(x) dx + C |
| 3. SOLVE FOR C: Apply initial condition (x_0, y_0) IMMEDIATELY.         |
| 4. ISOLATE y: Use algebra/exponentials to write y explicitly as f(x).   |
+-------------------------------------------------------------------------+

Step-by-Step Worked Example 1: Initial Value Problem

Problem: Solve the initial value problem $\frac{dy}{dx} = \frac{2x}{y}$ with initial condition $y(0) = -3$.

Step 1: Separate variables. Multiply both sides by $y , dx$: $y \, dy = 2x \, dx$

Step 2: Integrate both sides. $\int y \, dy = \int 2x \, dx \implies \frac{1}{2}y^2 = x^2 + C$

Step 3: Solve for $C$ immediately using $y(0) = -3$. $\frac{1}{2}(-3)^2 = (0)^2 + C \implies \frac{9}{2} = C$ $\frac{1}{2}y^2 = x^2 + \frac{9}{2}$

Step 4: Isolate $y$ explicitly. Multiply by $2$: $y^2 = 2x^2 + 9 \implies y = \pm \sqrt{2x^2 + 9}$

Selecting the correct sign: Because the initial condition requires $y(0) = -3 < 0$, we must select the negative square root:

$y(x) = -\sqrt{2x^2 + 9}$

(Warning: Selecting $+\sqrt{2x^2 + 9}$ yields $y(0) = +3 \neq -3$, violating the initial condition!)

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3. The Law of Natural Growth and Decay ($\frac{dy}{dt} = ky$)

The Fundamental Principle

In many natural phenomena, the rate of growth or decay of a quantity $y$ is directly proportional to the current amount present:

$\frac{dy}{dt} = k y$

• If $k > 0$: Exponential Growth (population growth, continuously compounded interest).
• If $k < 0$: Exponential Decay (radioactive decay, chemical decomposition).

Full Derivation of the Universal Solution

Step 1: Separate variables. $\frac{1}{y} \, dy = k \, dt$

Step 2: Integrate both sides. $\int \frac{1}{y} \, dy = \int k \, dt \implies \ln|y| = kt + C_1$

Step 3: Exponentiate both sides. $|y| = e^{kt + C_1} = e^{C_1} \cdot e^{kt} \implies y(t) = A e^{kt} \quad (\text{where } A = \pm e^{C_1})$

Step 4: Apply initial condition $y(0) = y_0$. $y(0) = A e^{k(0)} = A(1) = y_0 \implies A = y_0$

$\mathbf{y(t) = y_0 e^{kt}}$

   Exponential Growth (k > 0)                 Exponential Decay (k < 0)
   y ^                                        y ^
     |          .-'                             |  y_0
     |        .'                                |  *--.
     |       /                                  |  |   `.
     |     .'                                   |  |     `.
     |   .'                                     |  |       `-.
   y0| -*                                       |  |          `--.
     |                                          |  |              `--.
   --+-------------------> t                  --+--+-------------------> t
     0                                          0  t_half

Doubling Time ($k > 0$)

The time $t_d$ required for an initial population $y_0$ to double ($y = 2y_0$):

$2y_0 = y_0 e^{k t_d} \implies 2 = e^{k t_d} \implies \ln(2) = k t_d \implies \mathbf{t_d = \frac{\ln 2}{k}}$

Half-Life ($k < 0$ or $\frac{dy}{dt} = -ky$)

The time $t_h$ required for half of a radioactive substance to decay ($y = \frac{1}{2}y_0$):

$\frac{1}{2}y_0 = y_0 e^{-k t_h} \implies \frac{1}{2} = e^{-k t_h} \implies \ln(1/2) = -k t_h \implies -\ln 2 = -k t_h \implies \mathbf{t_h = \frac{\ln 2}{k}}$

Key Constant Property: Doubling time and half-life depend strictly on the rate constant $k$ and are completely independent of the initial amount $y_0$.

Step-by-Step Worked Example 2: Radioactive Decay & Radiocarbon Dating

Problem: Carbon-14 has a half-life of $5{,}730$ years. An ancient wooden artifact found at an archaeological site contains $25%$ of the Carbon-14 found in living trees. Determine the age of the artifact.

Step 1: Compute decay constant $k$. $k = \frac{\ln 2}{t_h} = \frac{\ln 2}{5730}$

Step 2: Set up decay equation with $y(t) = 0.25 y_0$. $0.25 y_0 = y_0 e^{-kt} \implies 0.25 = e^{-kt}$

Step 3: Solve for time $t$. $\ln(0.25) = -kt \implies \ln\left(\frac{1}{4}\right) = -kt \implies -\ln(4) = -kt$ $t = \frac{\ln 4}{k} = \frac{2\ln 2}{\frac{\ln 2}{5730}} = 2 \cdot 5730 = 11{,}460 \text{ years}$

(Inspection shortcut: $25% = \left(\frac{1}{2}\right)^2$, meaning exactly $2$ half-lives have elapsed: $2 \times 5{,}730 = 11{,}460$ years.)

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4. Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature $T(t)$ of an object is directly proportional to the difference between its temperature and the ambient temperature of the surrounding medium $T_{\text{env}}$:

$\frac{dT}{dt} = -k (T - T_{\text{env}}) \quad (k > 0)$

Derivation of Explicit Solution

1. Separate variables: $\frac{1}{T - T_{\text{env}}} , dT = -k , dt$
2. Integrate: $\ln|T - T_{\text{env}}| = -kt + C_1$
3. Exponentiate: $T - T_{\text{env}} = A e^{-kt}$
4. Apply initial condition $T(0) = T_0$: $T_0 - T_{\text{env}} = A$

$\mathbf{T(t) = T_{\text{env}} + (T_0 - T_{\text{env}}) e^{-kt}}$

Step-by-Step Worked Example 3: Cooling Coffee

Problem: A freshly brewed cup of coffee at temperature $90^\circ\text{C}$ is placed in a room maintained at a constant temperature of $20^\circ\text{C}$. After $10$ minutes, the coffee has cooled to $60^\circ\text{C}$. How many total minutes will it take for the coffee to reach $30^\circ\text{C}$?

Step 1: Set up the model parameters.

• $T_{\text{env}} = 20$
• $T_0 = 90 \implies T_0 - T_{\text{env}} = 90 - 20 = 70$
• $T(t) = 20 + 70 e^{-kt}$

Step 2: Find constant $k$ using $T(10) = 60$. $60 = 20 + 70 e^{-10k} \implies 40 = 70 e^{-10k} \implies e^{-10k} = \frac{4}{7}$ $-10k = \ln\left(\frac{4}{7}\right) \implies k = -\frac{1}{10}\ln\left(\frac{4}{7}\right) = \frac{1}{10}\ln\left(\frac{7}{4}\right) \approx 0.05596$

Step 3: Solve for $t$ when $T(t) = 30$. $30 = 20 + 70 e^{-kt} \implies 10 = 70 e^{-kt} \implies e^{-kt} = \frac{1}{7}$ $-kt = \ln\left(\frac{1}{7}\right) = -\ln(7) \implies t = \frac{\ln 7}{k}$ $t = \frac{\ln 7}{\frac{1}{10}\ln(7/4)} = 10 \cdot \frac{\ln 7}{\ln(1.75)} \approx 10 \cdot \frac{1.9459}{0.5596} \approx 34.77 \text{ minutes}$

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5. Slope Fields (Direction Fields)

A slope field (or direction field) is a graphical tool used to visualize the family of solutions to a first-order differential equation $\frac{dy}{dx} = f(x, y)$ without finding an explicit analytical antiderivative.

   Slope Field for dy/dx = y - 1
   y ^
   3 |  \  \  \  \  \  \  \  (slope m = 3 - 1 = +2)
   2 |  /  /  /  /  /  /  /  (slope m = 2 - 1 = +1)
   1 |  -  -  -  -  -  -  -  (slope m = 1 - 1 =  0) <--- Equilibrium Solution y = 1
   0 |  \  \  \  \  \  \  \  (slope m = 0 - 1 = -1)
  -1 |  /  /  /  /  /  /  /  (slope m = -1 - 1 = -2)
   --+-----------------------> x

Rules for Reading & Matching Slope Fields on the CLEP

1. Slopes depending ONLY on $x$ ($\frac{dy}{dx} = g(x)$):
   • Tangent segments are identical along any vertical column (same $x$).
   • Example: $\frac{dy}{dx} = 2x$. At $x = 0$, all slopes are horizontal ($m=0$). At $x = 1$, all slopes are $m=2$.
2. Slopes depending ONLY on $y$ ($\frac{dy}{dx} = h(y)$):
   • Tangent segments are identical along any horizontal row (same $y$).
   • Example: $\frac{dy}{dx} = y - 1$. Along the horizontal line $y = 1$, all slopes are $0$ (equilibrium solution).
3. Slopes depending on BOTH $x$ and $y$ ($\frac{dy}{dx} = f(x, y)$):
   • Check coordinate axes: What happens when $x = 0$ ($y$-axis) or $y = 0$ ($x$-axis)?
   • Check sign quadrants: Where is $\frac{dy}{dx} > 0$ (slanted up) vs $\frac{dy}{dx} < 0$ (slanted down)?
   • Example: $\frac{dy}{dx} = -\frac{x}{y}$. Slopes are $0$ when $x = 0$ (along $y$-axis) and vertical/undefined when $y = 0$ (along $x$-axis). The solution curves form concentric circles $x^2 + y^2 = C$.

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6. Comprehensive Differential Equations Comparison Table

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7. Common CLEP Pitfalls & Traps

1. The Denominator Differential Error: Never leave $dx$ or $dy$ in the denominator (e.g., writing $\frac{1}{dx} = \dots$). Differentials must always appear in the numerators of both sides before integrating.
2. The Constant Exponentiation Trap: When integrating $\frac{1}{y} , dy = k , dt$, the integration yields $\ln|y| = kt + C$. Exponentiating gives $y = e^{kt + C} = e^C e^{kt} = A e^{kt}$. Writing $y = e^{kt} + C$ is a critical algebraic error.
3. Ignoring Initial Condition Signs on Square Roots: When solving $y^2 = 2x^2 + C$, the explicit solution is $y = \pm \sqrt{2x^2 + C}$. You must explicitly choose $+$ or $-$ to match the sign of $y(x_0)$.
4. Confusing Natural Logarithm with Common Logarithm: Doubling time and half-life formulas require the natural logarithm $\ln(2) \approx 0.693$, never the base-10 logarithm $\log_{10}(2) \approx 0.301$.
5. Postponing Constant Evaluation: Always evaluate $C$ immediately after performing integration. Delaying $C$ until after isolating $y$ frequently introduces sign and power errors.