Differential Equations - Ordinary Differential Equations - First Order Linear Equations

where c is an arbitrary constant. If we have an initial condition such as F(ℓ) = mω2 (ℓ + r ), we find that ρ A0 ω2 (ℓ + r )2 c= + mω2 (ℓ + r ). 2

⊠

Example 3.2. A falling object attached to a parachute experiences a downward gravitational force of mg and an upward drag of mkv 2 , where g is the force due to gravity, k is a positive constant, and v(t) is its velocity at time t. OHP 11

mkv²

mg m

Thus we have dv m = mg − mkv 2 dt or

dv dv = dt. = g − kv 2 ⇔ 2 dt g − kv

Using partial fractions, this may be written as dv dv + = 2 dt. √ √ g − gkv g + gkv OHP 12

Setting ω =

√

gk, this leads to

1 [− ln(g − ωv) + ln(g + ωv)] = 2t + c˜ ω or

g + ωv = ce2ωt , g − ωv

c = eωc˜ .

Some algebra then yields g(ce2ωt − 1) . v(t) = 2ωt ω(ce + 1)

If we assume that v(0) = 0, we obtain c = 1 so that g(e2ωt − 1) v(t) = . ω(e2ωt + 1) As t → ∞, we see that v(t) → g/ω = terminal velocity.

√

g/k, the

OHP 13

⊠ §4 First order linear equations Recall that a first order linear o.d.e. is an equation of the form P0 (x)y ′ + P1 (x)y = G(x). Dividing by P0 (x) yields dy + p(x)y = g(x), (I.2) dx where p(x) = P1 (x)/P0 (x) and g(x) = G(x)/P0 (x). OHP 14

If g is the zero function, then the linear o.d.e. is said to be homogeneous. In the homogeneous case, the equation is separable and we have dy = − p(x) dx. y Solving this equation yields ln(y) = −P(x) + c, ˜ where R P(x) = p(x) dx and c˜ is some arbitrary constant.

This solution may be written as y(x) = ce−P(x) , where c = ec˜ is an arbitrary constant. All solutions have to be of this form and is called the general solution of the homogeneous equation. Theorem I.1. The general solution of the equation dy + p(x)y = 0 dx

is given by y(x) = ce−P(x) , where P(x) =

Z

p(x) dx

(I.3)

and c is an arbitrary constant. OHP 15