Lecture Notes in Differential Equations - Bruce E. Shapiro

191
This is a first order linear equation.An integrating factor is µ = e −3t . Multiplying
both sides of (22.27) by µ,
d (
ze
−3t ) = 3e −3t sin t (22.28)
dt
Integrating,
∫
ze −3t = 3
e −3t sin tdt = 3
10 e−3t (−3 sin t − cos t) + C (22.29)
or
z = − 3
10 (3 sin t + cos t) + Ce−3t (22.30)
Substituting back for z = y ′ − 7y from equation (22.25) gives us
y ′ − 7y = − 3 10 (3 sin t + cos t) + Ce3t (22.31)
This is also a first order linear ODE, which we know how to solve. An
integrating factor is µ = e −7t . Multiplying (22.31) by µ and integrating
over t gives
(y ′ − 7y)(e −7t ) =
[− 3 10 (3 sin t + cos t) + Ce3t ]
(e −7t ) (22.32)
d (
ye
−7t ) = − 3 dt
10 e−7t (3 sin t + cos t) + Ce −4t (22.33)
∫
ye −7t =
[− 3 ]
10 e−7t (3 sin t + cos t) + Ce −4t dt (22.34)
= − 3 ∫
∫
e −7t (3 sin t + cos t)dt + C e −4t dt (22.35)
10
Integrating the last term would put a −4 into the denominator; absorbing