More on Solutions - ODE Notes

1. Newton's Law of Cooling

Variables:

$ T $: Object's temperature
$ t $: Time
$ A $: Ambient temperature

The rate of change of temperature is proportional to the difference between ambient temperature and the object's temperature: \[ \frac{dT}{dt} \propto (A - T) \longrightarrow \frac{dT}{dt} = k(A - T) \] Where $ k > 0 $.

Observations:

If $ T < A $, then $ T' > 0 $ (The object warms up).
If $ T = A $, then $ T' = 0 $ (Thermal equilibrium).
If $ T > A $, then $ T' < 0 $ (The object cools).

2. General and Particular Solutions

If an Ordinary Differential Equation (ODE) is in the form $ y' = f(x) $, the general solution is found via integration: $ y = \int f(x) dx + C $.

Example 1: Basic Integration

Find the general solution for $ y' = 2x + 3 $:

\[ \int dy = \int (2x + 3) dx \] \[ y + c_1 = x^2 + 3x + c_2 \] \[ y = x^2 + 3x + C \quad \text{(General Solution)} \]

Find the particular solution given the initial value $ y(1) = 2 $:

\[ 2 = (1)^2 + 3(1) + C \] \[ 2 = 4 + C \implies C = -2 \] \[ y = x^2 + 3x - 2 \quad \text{(Particular Solution/Solution to IVP)} \]

Example 2: Inverse Trigonometric Form

Find the general solution for $ \{y' = \frac{12}{x^2 + 16}\} $:

Recall: $ \frac{d}{dx}[\tan^{-1}(\frac{x}{a})] = \frac{a}{x^2 + a^2} $

\[ \frac{dy}{dx} = 3 \cdot \frac{4}{x^2 + 4^2} \] \[ y = 3\tan^{-1}\left(\frac{x}{4}\right) + C \]

Find the particular solution for $ y(0) = 1 $:

\[ 1 = 3\tan^{-1}(0) + C \longrightarrow 1 = C \] \[ y = 3\tan^{-1}\left(\frac{x}{4}\right) + 1 \]

3. Acceleration, Velocity, and Position

For constant acceleration $ a(t) = x''(t) = 7 $:

Integrate once for velocity: $ v(t) = x'(t) = 7t + C_1 $
Integrate again for position: $ x(t) = \frac{7}{2}t^2 + C_1t + C_2 $

Note: A 2nd order ODE general solution contains two unknown constants. These correspond to physical initial conditions:

$ C_1 = v(0) = v_0 $ (Initial velocity)
$ C_2 = x(0) = x_0 $ (Initial position)

Final form: $ x(t) = \frac{7}{2}t^2 + v_ot + x_o $

4. Modeling an ODE from Geometric Properties

Problem: Write an ODE for $ y(x) $ such that the tangent line at $ (x, y) $ passes through the point $ (-y, x) $.

$An xy-plane with a line drawn through two points $ (x,y) $ and $ (-y,x) $.$

The slope of the tangent line $ y' $ is determined by the two points $ (x, y) $ and $ (-y, x) $:

\[ \text{slope} = \frac{\Delta y}{\Delta x} = \frac{y - x}{x - (-y)} = \frac{y - x}{x + y} \]

Therefore, the ODE is:

\[ \frac{dy}{dx} = \frac{y - x}{x + y} \]