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PAGE 1

1.1 (Continued)

Direction / Slope Field

From last time: \( y' = y \). We graph the slope at different \( (t, y) \).

A coordinate system with t and y axes showing a slope field where slopes are constant horizontally.

\( y' = y \) does not depend on \( t \)

Slopes don't change left/right

Let's try \( y' = y(3-y) \)

First, notice \( y' = 0 \) at \( y = 0, y = 3 \) (horizontal slopes).

Slope field for y' = y(3-y) showing horizontal slopes at y=0 and y=3, with integral curves.

Again, no \( t \), so the same slopes on each horizontal line.

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Now we check the regions \( y < 0, \quad 0 < y < 3, \quad y > 3 \)

For \( y < 0 \)

\[ y' = y(3-y) < 0 \]

(\(- \) times \( + \))

  • All negative slopes
  • Near \( y = 0 \), shallow negative slope
  • As \( y \to -\infty, y' \to -\infty \) (as we go down, greater negative slopes)

For \( 0 < y < 3 \)

\[ y' = y(3-y) > 0 \]

(\(+ \) times \( + \))

Starts nearly 0, increases as \( y \) increases, then decreases to 0 at \( y = 3 \).

For \( y > 3 \)

\[ y' = y(3-y) < 0 \]

(\(+ \) times \( - \))

  • As \( y \to \infty, y' \to -\infty \)
  • Steeper as we go up.
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Analysis of Initial Conditions

  • If \( y(0) < 0 \) (initial condition) \( t \to \infty, y \to -\infty \)
  • If \( 0 < y(0) < 3 \) \( t \to \infty, y \to 3 \)
  • If \( y(0) > 3, \) \( t \to \infty, y \to 3 \)

Example: Analyzing a Non-Autonomous Equation

Let's try \( y' = -2 + t - y \)

Slopes depend on both \( t \) and \( y \)

Efficient way to analyze: find out where \( y' = 0 \)

\[ 0 = -2 + t - y \]

\[ y = t - 2 \]

\( \to \) on this line slopes \( (y') \) are zero

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Slope field on t-y axes with red slope segments. A solution curve approaches the line y = t-2.

Slope Field Analysis

\[ y' = (t - 2) - y \]

Region: Above \( t - 2 \)

\( y > t - 2 \)

\( y' < 0 \)

More above, more negative

Region: Below \( t - 2 \)

\( y < t - 2 \)

\( y' > 0 \)

More positive below

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1.2 Solutions of Some Differential Eqs.

\[\frac{dy}{dt} = ay - b \quad a, b \text{ constants}\]

Shows up in free fall:

Free body diagram showing upward air resistance force \gamma v and downward gravity force mg.

Newton's 2nd Law: \(F = ma\)

\(v\): velocity then \(a = \frac{dv}{dt}\)

\[mg - \gamma v = m \frac{dv}{dt}\]

rewrite:

\[\frac{dv}{dt} = g - \frac{\gamma}{m} v\]
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let's solve one:

\[\frac{dy}{dt} = -3y + 10\]

can't solve by integrating both sides directly (we don't know \(y\))

but we can do this:

\[\frac{dy}{dt} = -3 \left( y - \frac{10}{3} \right)\]

divide by \(y - \frac{10}{3} \quad (y \neq \frac{10}{3})\)

\[\frac{1}{y - \frac{10}{3}} \frac{dy}{dt} = -3\]

let \(u = y - \frac{10}{3}\) then \(\frac{du}{dt} = \frac{dy}{dt}\)

\[\frac{1}{u} \frac{du}{dt} = -3\]
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Solving the Differential Equation

Integrate both sides with respect to \( t \):

\[ \int \frac{1}{u} \frac{du}{dt} dt = \int -3 dt \]
\[ \int \frac{1}{u} du = \int -3 dt \]
\[ \ln |u| = -3t + C \]

Substituting back for \( u \):

\[ \ln \left| y - \frac{10}{3} \right| = -3t + C \]

Exponentiating both sides:

\[ \left| y - \frac{10}{3} \right| = e^{-3t + C} = e^{-3t} \cdot e^C \]
\[ y - \frac{10}{3} = \pm e^C \cdot e^{-3t} = C e^{-3t} \]

Note: We call the constant \( \pm e^C \) simply \( C \).

\[ y = \frac{10}{3} + C e^{-3t} \]

General Solution

It is a collection of infinitely-many solutions (one for each value of \( C \)).

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Finding the Particular Solution

To find \( C \), we need one point on the curve.

Typically, \( y(0) = y_0 \) (initial condition). This makes the problem an Initial-Value Problem (IVP).

\[ y = \frac{10}{3} + C e^{-3t} \]

Using the initial condition \( y(0) = y_0 \), sub in \( y = y_0 \) and \( t = 0 \):

\[ y_0 = \frac{10}{3} + C \quad \text{so,} \quad C = y_0 - \frac{10}{3} \]

Then:

\[ y(t) = \frac{10}{3} + \left( y_0 - \frac{10}{3} \right) e^{-3t} \]

Particular Solution