Lecture Notes in Differential Equations - Bruce E. Shapiro

27
So far any function µ will work, but not any function will help. We want
to find an particular function µ such that the left hand side of the equation
becomes
d(µy)
= µy ′ + µpy = µq (4.14)
dt
The reason for looking for this kind of µ is that if we can find µ then
d
(µy) = µq (4.15)
dt
Multiplying both sides by dt and integrating gives
∫
∫
d
dt (µ(t)y)dt = µ(t)q(t)dt (4.16)
Since the integral of an exact derivative is the function itself,
∫
µ(t)y = µ(t)q(t)dt + C (4.17)
hence dividing by µ, we find that if we can find µ to satisfy equation
4.14 then the general solution of equation 4.10 is
y = 1 [∫
]
µ(t)q(t)dt + C
(4.18)
µ(t)
So now we need to figure out what function µ(t) will work.
product rule for derivatives
From the
d
dt (µy) = µy′ + µ ′ y (4.19)
Comparing equations 4.14 and 4.19,
µy ′ + µ ′ y = µy ′ + µpy (4.20)
µ ′ y = µpy (4.21)
µ ′ = µp (4.22)
Writing µ ′ = dµ/dt we find that we can rearrange and integrate both sides
of the equation:
∫
dµ
= µp (4.23)
dt
dµ
= pdt (4.24)
µ
∫ 1
µ dµ = pdt (4.25)
∫
ln µ = pdt + C (4.26)