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1.2 Solutions of Some DEs

Object in free fall

\[ v(t) \rightarrow \text{velocity } \left( \frac{dy}{dt} \right) \]

Free body diagram showing a central point with an upward arrow labeled c dy/dt (drag) and a downward arrow labeled mg (weight).

DE:

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

Let \( m = 10 \text{ kg} \) and \( c = 2 \text{ kg/s} \)

\[ 10 \frac{dv}{dt} = 98 - 2v \]\[ \frac{dv}{dt} = 9.8 - 0.2v \quad \Rightarrow \quad v(t) = ? \]

Assume: \( v(0) = 0 \)

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\[ \frac{dv}{dt} = 0.2(49 - v) = -0.2(v - 49) \]\[ \frac{1}{v - 49} \frac{dv}{dt} = -0.2 \]

Recall from calculus:

\[ \frac{d}{dt} \ln |v(t) - 49| = rac{1}{v(t) - 49} \frac{dv}{dt} \]
\[ \frac{d}{dt} \ln |v - 49| = -0.2 \]

Integrate with respect to \( t \)

\[ \ln |v - 49| = -0.2t + A \quad \text{(A is a constant)} \]\[ e^{\ln |v - 49|} = e^{-0.2t + A} \]\[ v - 49 = e^{-0.2t} e^A \]

Let \( e^A = C \) (also a constant).

\[ v - 49 = Ce^{-0.2t} \]\[ v = 49 + Ce^{-0.2t} \]

Find \( C \) using initial condition \( v(0) = 0 \):

\[ 0 = 49 + C \quad \text{so} \quad C = -49 \]
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So

\[ V(t) = 49 - 49e^{-0.2t} \]

note: as \( t \to \infty \), \( v \to 49 \) (because \( \lim_{t \to \infty} e^{-t} = 0 \))

terminal velocity

(weight balances drag)

\[ = \frac{mg}{c} \]
Graph of velocity v versus time t showing an exponential approach from zero to a horizontal asymptote at v=49.

now, assume \( V(0) = V_0 \)

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\[ V(t) = 49 + Ce^{-0.2t} \quad V(0) = V_0 \] \[ V_0 = 49 + C \quad \text{so} \quad C = V_0 - 49 \]
\[ V(t) = 49 + (V_0 - 49)e^{-0.2t} \]
Graph showing two velocity curves approaching a terminal velocity of 49: one decreasing from V0 and one increasing from 0.
  • if \( V(0) > \text{terminal} \), drag \( > \) weight initially, so lose speed until terminal
  • if \( V(0) < \text{terminal} \), weight \( > \) drag initially, so gain speed until terminal
  • if \( V(0) = \text{terminal} \), then drag \( = \) weight, stays at terminal
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We can use the same steps to solve all equations of the form

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

\( a, b \) constants

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1.3 Classifications of DEs

Ordinary DE (ODE)

Equations with ordinary derivatives (no partial derivs).

Partial DE (PDE)

Equations with at least one partial deriv.

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

Order of DE

Order is the order of the highest derivative in eq.

\( y'' + y' + y = 0 \)ODE of 2nd order
\( \alpha^2 u_{xx} = u_t \)PDE of 1st order
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Linear vs Nonlinear Differential Equations

A Differential Equation (DE) is linear if it is a linear function of all its derivatives of the dependent variable.

Criteria for Linearity:

  • Not multiplied together
  • Not raised to any power other than 1
  • Not part of transcendental functions (e.g., exponential \( e^y \), sine, cosine, etc.)

Examples

\[ y'' + y' + y = 0 \]
linear "easy" to solve
\[ y'' + (y')^2 + y = 0 \]
nonlinear "hard to solve
\[ y'' + y'y + \sin(y) = 0 \]
nonlinear
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\[ y'' + y = 0 \]

linear, 2nd order ODE

\[ t y'' + y = 0 \]

\( t \): independent variable

linear, 2nd order ODE

\[ t^2 y'' + t y' = \sin(t) \]

transcendental but w/ indep. var.

linear 2nd order ODE

\[ \frac{d^2\theta}{dt^2} + \frac{g}{L} \sin(\theta) = 0 \]

\( \theta \): dependent variable

nonlinear

Diagram of a swinging pendulum of length L at an angle \theta from the vertical.
Swinging pendulum model
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Solution: any function that satisfies the DE

example:

\[ \frac{dy}{dt} = y \]

function = its own deriv.

\( y = e^t \)    \( y' = e^t \)

\( \rightarrow y = Ce^t \)

solution of DE

many DEs will have \( y = e^{rt} \) as a solution

\( r = \text{constant} \)

example:

\[ y'' + 2y' - 3y = 0 \]

for what values of \( r \) is \( y = e^{rt} \) a solution?

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plug \( y = e^{rt} \) into \( y'' + 2y' - 3y = 0 \)

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

\( y'' = r^2e^{rt} \)

\[ r^2e^{rt} + 2re^{rt} - 3e^{rt} = 0 \]\[ e^{rt}(r^2 + 2r - 3) = 0 \]

since \( e^{rt} \neq 0 \),    \( r^2 + 2r - 3 = 0 \)

\( (r + 3)(r - 1) = 0 \)

\( r = -3, \text{ or } r = 1 \)

\( y'' + 2y' - 3y = 0 \) has \( e^{-3t} \) and \( e^t \) as solutions