Unit 7 Study Notes: Understanding Differential Equations

Introduction to Modeling Change

Calculus is fundamentally the study of change. In previous units, you learned that the derivative, $\frac{dy}{dx}$, represents the instantaneous rate of change of a quantity. In Unit 7, we reverse this process. Instead of starting with a function and finding its derivative, we start with an equation involving a derivative—a Differential Equation—and work backward to find the original function.

What is a Differential Equation?

A differential equation (DE) is an equation that relates a function $y$, its independent variable $x$, and one or more of its derivatives (like $y'$ or $y''$).

  • Goal: To find a function $y = f(x)$ that makes the equation true.
  • General Solution: A family of functions containing an arbitrary constant $C$ (e.g., $y = x^2 + C$). This represents all possible curves that satisfy the rate of change.
  • Particular Solution: A specific function (e.g., $y = x^2 + 5$) found by using an initial condition (a known point $(x, y)$) to solve for $C$.

Verifying Solutions

Before you learn to solve them from scratch, you must know how to check if a function is a valid solution to a differential equation.

The Method

  1. Differentiate: Take the derivative of the proposed solution $y$.
  2. Substitute: Plug $y$ and $\frac{dy}{dx}$ into the given differential equation.
  3. Verify: Check if the left-hand side (LHS) equals the right-hand side (RHS).

Example:
Is $y = e^{-3x}$ a solution to the differential equation $y' + 3y = 0$?

  • Step 1: Find $y'$. Using the chain rule, $y' = -3e^{-3x}$.
  • Step 2: Substitute into the DE.
    (-3e^{-3x}) + 3(e^{-3x}) = 0
  • Step 3: Simplify.
    -3e^{-3x} + 3e^{-3x} = 0 \quad \checkmark
    Since $0=0$, it is a solution.

Slope Fields

Algebraic solutions aren't the only way to understand differential equations. Slope Fields provide a geometric (visual) representation of the general solution.

What is a Slope Field?

A slope field is a graph composed of many small line segments drawn at coordinate grid points. At any specific point $(x, y)$, the derivative $\frac{dy}{dx}$ gives you the slope of the tangent line at that point.

Slope field showing tangent lines flow

Constructing a Slope Field

To draw a slope field manually:

  1. Pick a point on the grid, say $(1, 2)$.
  2. Plug these coordinates into the differential equation to find the slope value.
  3. Draw a short line segment at $(1, 2)$ with that specific slope.
  4. Repeat for other points.

Tip: Look for patterns to save time.

  • If the DE is $\frac{dy}{dx} = x$ (depends only on $x$), the slopes will be identical in every vertical column.
  • If the DE is $\frac{dy}{dx} = y$ (depends only on $y$), the slopes will be identical in every horizontal row.

Sketching Solution Curves

The AP Exam often asks you to sketch a particular solution curve passing through a given initial point.

The Strategy: "Go with the Flow"

  1. Mark the Start: Plot the given initial condition point (e.g., $(0, 1)$).
  2. Follow the Lines: Draw a curve that moves significantly parallel to the nearby slope segments. Think of the slope segments as currents in a river; your line should flow with them.
  3. Boundaries: Your curve should extend to the edges of the graph unless it hits an asymptote or a place where the derivative is undefined.

Example of a solution curve drawn through a slope field

Solving by Separation of Variables

This is the most critical algebraic skill in Unit 7. When given a first-order differential equation, you must rearrange it so all $y$ terms are on one side and all $x$ terms are on the other before integrating.

The SIPPY Method

A mnemonic to remember the steps for finding a particular solution is SIPPY:

  1. S - Separate: Multiply/divide to get $dy$ with $y$ terms on the left, and $dx$ with $x$ terms on the right.
  2. I - Integrate: Apply $\int$ to both sides.
  3. P - Plus C: Immediately add $+C$ to the $x$-side (right side).
  4. P - Plug In: Substitute the initial condition $(x, y)$ to solve for $C$.
  5. Y - Y equals: Algebraically manipulate the equation to isolate $y$.

Worked Example

Problem: Find the particular solution to $\frac{dy}{dx} = \frac{4x}{y}$ with initial condition $y(0) = 5$.

1. Separate:
multiply both sides by $y$ and by $dx$:
y \, dy = 4x \, dx

2. Integrate:
\int y \, dy = \int 4x \, dx
\frac{y^2}{2} = 2x^2

3. Plus C:
Do not forget this step!
\frac{y^2}{2} = 2x^2 + C

4. Plug In:
Use $x=0, y=5$ to find $C$:
\frac{5^2}{2} = 2(0)^2 + C
\frac{25}{2} = 0 + C \quad \Rightarrow \quad C = \frac{25}{2}
Update equation: $\frac{y^2}{2} = 2x^2 + \frac{25}{2}$

5. Y equals:
Isolate $y$. First, multiply by 2:
y^2 = 4x^2 + 25
Take the square root:
y = \pm\sqrt{4x^2 + 25}
Crucial Decision: Since our initial $y$ was positive ($5$), we keep the positive root.

Final Solution:
y = \sqrt{4x^2 + 25}

Exponential Models

A specific type of differential equation appears frequently in real-world modeling (population growth, radioactive decay).

The Law of Exponential Change

If the rate of change of a quantity $y$ is proportional to the quantity itself, we write:
\frac{dy}{dt} = ky

Where $k$ is the constant of proportionality.

  • If $k > 0$: Exponential Growth (Population, Bacteria)
  • If $k < 0$: Exponential Decay (Radioactive isotopes)

The General Solution

You can solve this using Separation of Variables, but you should memorize the result:
y(t) = y_0 e^{kt}

  • $y(t)$ = amount at time $t$
  • $y_0$ = initial amount (at $t=0$)

Newton's Law of Cooling

Sometimes rate of change depends on the difference between the object's temperature ($T$) and the surrounding room temperature ($Ts$). \frac{dT}{dt} = k(T - Ts)

To solve this, treat $(T - T_s)$ as a single variable during separation:

  1. $\frac{1}{T-T_s} dT = k \, dt$
  2. Integrate to get $\ln|T - T_s| = kt + C$
  3. Exponentiate both sides to solve for $T$.

Common Mistakes & Pitfalls

  1. The Missing "+C":

    • Mistake: Forgetting to write $+C$ immediately after integrating.
    • Correction: If you add $C$ at the very end, your algebra will be wrong (e.g., $e^{x+C}$ is very different from $e^x + C$).

  2. Bad Separation Algebra:

    • Mistake: If $\frac{dy}{dx} = x + y$, students often try to divide by $y$ to get $\frac{1}{y}dy = x dx$. This is illegal. You can only separate by multiplication or division.
    • Note: $\frac{dy}{dx} = x+y$ cannot be solved by separation of variables.

  3. Ignoring Absolute Values with Logarithms:

    • Mistake: Integrating $\int \frac{1}{y} dy = \ln(y)$.
    • Correction: The integral is $\ln|y| + C$. The absolute value matters for determining the domain and the correct branch of the solution.

  4. Slope Field Sloppiness:

    • Mistake: Drawing a solution curve that crosses through a region where the slope is undefined, or crossing over its own asymptote.
    • Correction: Solution curves are continuous functions. They cannot jump over asymptotes.

  5. Picking the Wrong Root:

    • Mistake: Leaving the answer as $y = \pm\sqrt{…}$ or arbitrarily picking positive.
    • Correction: Check your initial condition. If $y(0) = -3$, your solution must involve the negative root ($y = -\sqrt{…}$).