Differential Equations - Ordinary Differential Equations - Integrating Factors

§5 Integrating factors To obtain the general solution of a first order linear o.d.e., we could solve the homogeneous equation and than find a particular solution by guessing. Here we obtain the general solution of a first order linear o.d.e. by the use of an integrating factor. This method yields the solution of the homogeneous equation and a particular solution ‘in one go’. This method does not extend to higher order linear o.d.e.’s. Recall that the first order linear o.d.e. is dy + p(x)y = g(x). dx Now define the integrating factor ϕ(x) by ϕ(x) := e

R

p(x) dx

which has derivative ϕ ′ (x) = p(x)e

R

p(x) dx

= p(x)ϕ(x). OHP 21

By multiplying the o.d.e. by ϕ(x), we obtain ϕ(x)

dy + ϕ(x) p(x)y = ϕ(x)g(x). dx

(I.5)

Note that d (ϕ(x)y) = ϕ(x)y ′ +ϕ ′ (x)y = ϕ(x)y ′ +ϕ(x) p(x)y dx is the left-hand-side of (I.5). Thus d (ϕ(x)y) = ϕ(x)g(x). dx Integration with respect to x then yields Z ϕ(x)y = ϕ(x)g(x) dx + c and so y(x) =

Z

ϕ(x)g(x) dx + c ϕ(x)

.

This gives the general solution of the first order linear o.d.e. (I.2). This technique requires two integrations. In terms of Theorem I.2, c/ϕ(x) corresponds to ce−P(x) R and ϕ(x)g(x) dx/ϕ(x) is a particular solution y p (x). OHP 22

Example 5.1. For the electrical circuit of Example 4.1, suppose that V (t) = V0 sin(ωt). From (I.4) we have dI V0 sin(ωt) + αI = , dt L where α = R/L. The integrating factor is ϕ(t) = e Thus eαt or

R

α dt

= eαt .

V0 sin(ωt) dI + αeαt I = eαt , dt L d αt
αt V0 sin(ωt) . e I =e dt L

Integrating both sides yields V0 eαt e I (t) = [α sin(ωt) − ω cos(ωt)] + c. × 2 2 L α +ω αt

OHP 23

If I (0) = 0, we obtain V0 (α sin(ωt) I (t) = L(α 2 + ω2 ) V0 ω −αt − ω cos(ωt)) + e . 2 2 L(α + ω ) If β = tan−1 (ω/α), then sin(β) = √

ω α2

+ ω2

and

cos(β) = √

α α2

+ ω2

.

Also, sin(ωt − β) = sin(ωt) cos(β) − sin(β) cos(ωt) α ω cos(ωt) = sin(ωt) √ −√ 2 2 2 2 α +ω α +ω so we have V0

V0 ω −αt I (t) = √ e . sin(ωt − β) + 2 + ω2 ) 2 2 L(α L α +ω OHP 24