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1.1 Intro to Differential Equations

differential equation: an equation that contains a derivative

for example,

\[ \frac{dy}{dx} = f(x) \quad \text{or} \quad y' = f(x) \]

e.g.

\[ \frac{dy}{dx} = \cos x \rightarrow y = \sin x + C \quad \text{(Calculus!)} \]

Another Example

\[ \frac{d^2r}{dt^2} = - \frac{GM}{r^2} \]

Newton's Law of Gravitation

goal: find \( r(t) \)

One More: Logistic Growth

\[ \frac{dP}{dt} = k(L - P) \]
  • \( P(t) \) is population
  • \( L \) : pop. capacity

goal: find solution

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Order of Differential Equation

order of differential eq: highest derivative

\[ y' = y \]

1st-order

\[ r'' = - \frac{GM}{r^2} \]

2nd-order

Solution

solution: a function that satisfies the diff. eq.

for example,

\[ y' = y \]

solution: \( y = ? \)

\( y \) is its own derivative \( \rightarrow e^x \)

in fact, any multiple

so, \( y = Ce^x \)

we can verify that: it must satisfy \( y' = y \)

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Differential Equation Solutions

Does \( y = Ce^x \) satisfy \( y' = y \)?

\( y' = Ce^x \) (Left)

\( y = Ce^x \) (Right)

}
yes.

Solution Curves

Each \( C \) defines a solution curve.

Coordinate graph with x and y axes showing three exponential curves: two positive and one negative.

Example

Is \( y = \ln(x + c) \) a solution to \( e^y y' = 1 \)?

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\( e^y y' = 1 \)

\( y = \ln(x + c) \)

\( y' = \frac{1}{x + c} \)

\( e^y = e^{\ln(x + c)} = x + c \)

LHS: \( e^y y' = (x + c) \cdot \frac{1}{x + c} = 1 \)

RHS: \( 1 \)

So, \( y = \ln(x + c) \) is a solution.

Follow-up Question

Find \( C \) such that the solution goes through \( (0, 0) \).

\( y = \ln(x + c) \)

\( 0 = \ln(0 + c) = \ln(c) \quad \text{so} \quad C = 1 \)

Each \( C \) defines a new curve based on \( y(0) = \text{something} \)

Initial condition
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Example: Solving a Second-Order Linear Differential Equation

Given the differential equation:

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

Find all values of \( r \) such that \( y = e^{rx} \) is a solution.

Again, if \( y = e^{rx} \) is a solution, then \( y'' + y' - 2y = 0 \). We calculate the derivatives:

\[ \begin{aligned} y &= e^{rx} \\ y' &= re^{rx} \\ y'' &= r^2e^{rx} \end{aligned} \implies y'' + y' - 2y = 0 \]

Substituting these into the original equation:

\[ r^2e^{rx} + re^{rx} - 2e^{rx} = 0 \quad \text{for what } r? \]

Factoring out the exponential term:

\[ (e^{rx})(r^2 + r - 2) = 0 \]

Since \( e^{\text{anything}} \neq 0 \), we must have:

\[ r^2 + r - 2 = 0 \implies (r + 2)(r - 1) = 0 \]
\[ r = -2 \text{ or } r = 1 \]

So, \( e^x \) and \( e^{-2x} \) are solutions.

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Example: Deriving a Differential Equation from Geometric Properties

Write a differential equation \( \frac{dy}{dx} = f(x, y) \) such that the line tangent to the solution \( y = g(x) \) at \( (x, y) \) goes through \( (\frac{x}{2}, 0) \).

Coordinate graph showing a curve y=g(x) and its tangent line at (x,y) intersecting the x-axis at (x/2, 0).

\( \frac{dy}{dx} = ? \) that describes what we see on the left.

Derivation

\( \frac{dy}{dx} \) is the slope of the line tangent to \( y \).

The tangent line goes through the points \( (x, y) \) and \( (\frac{x}{2}, 0) \).

So the slope is:

\[ \text{slope} = \frac{y - 0}{x - \frac{x}{2}} = \frac{y}{\frac{x}{2}} = \frac{2y}{x} \]

So the differential equation is:

\[ \frac{dy}{dx} = \frac{2y}{x} \]