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PAGE 1

1.2 (continued)

last time we solved

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

Solution: \( y(t) = \frac{10}{3} + (y_0 - \frac{10}{3})e^{-3t} \) \( y(0) = y_0 \)

let's interpret this in terms of free-fall problem

Free-body diagram showing a particle with an upward force vector labeled \gamma v and a downward force vector labeled mg.
\[ m\frac{dv}{dt} = mg - \gamma v \]\[ \frac{dv}{dt} = g - \frac{\gamma}{m}v \]

\( \frac{dy}{dt} = -3y + 10 \rightarrow g = 10 \), \( \frac{\gamma}{m} = 3 \rightarrow \) air resistance is 3 times the velocity

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Slope field for \( \frac{dy}{dt} = -3y + 10 \)

Slope field graph on y vs t axes showing solution curves converging to a horizontal asymptote at y = 10/3.

\( y' = 0 \rightarrow y = \frac{10}{3} \)

if \( y < \frac{10}{3}, y' > 0 \)

\( y > \frac{10}{3}, y' < 0 \)

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

\( \frac{dv}{dt} = 0 \rightarrow \) gravity = air resistance

object falls w/ constant velocity when that happens \( \rightarrow \) terminal velocity

(low when \( \gamma \) is high, parachute, for example)

here, terminal velocity is \( \frac{10}{3} \)

  • if \( y(0) < \frac{10}{3} \), initially weight > resistance \( \rightarrow \) speed up until weight = resistance \( \rightarrow \) terminal velocity
  • if \( y(0) > \frac{10}{3} \), resistance dominates until gravity catches up
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Solution tells us the same

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

For \( t > 0 \), \( e^{-3t} > 0 \) and decreasing

  • If \( y_0 > \frac{10}{3} \), then \( y > \frac{10}{3} \) but decreases until \( \frac{10}{3} \)
  • If \( y_0 < \frac{10}{3} \), then \( y < \frac{10}{3} \) but increases until \( \frac{10}{3} \)

Summary:

We can solve

\[ \frac{dy}{dt} = ay - b \]

Other interpretation

\[ \frac{dy}{dt} = 2y - 1 \]

Population grows at rate twice of its size but we have a constant loss of 1

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1.3 Classification of Differential Eqs.

Two major families:

  • Ordinary diff. eqs. (ODE)
  • Partial diff. eqs. (PDE)

ODE: contains only ordinary derivatives (MA 366)

PDE: can contain partial derivatives (MA 303)

\[ \frac{dv}{dt} = g - \frac{\gamma}{m}v \](ODE)free fall
\[ \frac{dP}{dt} = k(L - P) \]logistic growth(ODE)
\[ m\frac{d^2y}{dt^2} + c\frac{dy}{dt} + ky = \cos(t) \]mass-spring-damper(ODE)

Dependent variables \( (v, P, y) \) only depend on one independent variable \( (t) \)

\( \rightarrow \) deriv. always with respect to \( t \) (ordinary)

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\[ \alpha^2 \frac{\partial^2 u(x,t)}{\partial x^2} = \frac{\partial u(x,t)}{\partial t} \] heat equation (PDE)

\[ a^2 \frac{\partial^2 u(x,t)}{\partial x^2} = \frac{\partial^2 u(x,t)}{\partial t^2} \] wave equation (PDE)

\( u \) depends on multiple independent variables \( (x, t) \)
\( \rightarrow \) partial derivatives

Order of a Differential Equation

The order of a diff. eq. is the order of the highest derivative.

\[ v' = g - \frac{\gamma}{m} v \] 1st-order

\[ y''' + 5y'' - 10y' + y = 0 \] 3rd-order

heat and wave eqs above are both 2nd-order

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Linearity

if a diff. eq. can be written in the form of

\[ F(t, y, y', y'', \dots, y^{(n)}) = 0 \text{ and } F \text{ is linear function} \]

then the differential equation is said to be linear

else it's nonlinear

Examples

for example,

\[ 5y'' + 3y' + 2y = t \]\[ 5y'' + 3y' + 2y - t = 0 \]

linear because none of the coefficients of \( y \) and its derivatives contain the dependent variable \( (y) \)

compare to

\[ y'' + y^2 = 0 \]\[ y'' + y \cdot y = 0 \]

\( \leftarrow \) coefficient of \( y \) has \( y \rightarrow \) nonlinear

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\[ ty' + 3 = 0 \]

  • ↑ coefficients of \( y, y', y'', \text{etc} \) don't contain any \( y \)
  • → linear

\[ y'' + \sin(y) = 0 \]

nonlinear because \( \sin(y) \) is NOT a linear function of \( y \)

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

nonlinear because \( \frac{1}{r^2} \) is not a linear function of \( r \)

Generally, nonlinear eqs. are harder to solve

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A solution is a function that satisfies the diff. eq.

for example, \( y = e^t \) is a solution to \( y' = y \)

because

\[ \underbrace{\frac{d}{dt}(e^t)}_{\frac{d}{dt}y} = \underbrace{e^t}_{y} \]

for example, \( y = \cos(t) \) and \( y = \sin(t) \) are solutions to

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

can verify by plugging into the diff. eq.

generally, an \( n^{\text{th}} \)-order eq. has \( n \) solutions

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One type of eq. we study a lot is linear constant-coeff diff. eq.

Solutions are in the form of \( y = e^{rt} \)

r: constant

For example, \( y'' + y' - 6y = 0 \)

sub \( y = e^{rt} \)

\( y' = re^{rt} \)

\( y'' = r^2 e^{rt} \) into diff. eq.

\[ r^2 e^{rt} + re^{rt} - 6e^{rt} = 0 \]

divide by \( e^{rt} \) since \( e^{rt} \neq 0 \) for any \( r, t \)

\[ (r^2 + r - 6) = 0 \] \[ (r + 3)(r - 2) = 0 \]

\( r = -3, r = 2 \)

Solutions:

\( y = e^{-3t} \)

\( y = e^{2t} \)