Lecture Notes in Differential Equations - Bruce E. Shapiro

31
Since p(t) = 1 (the coefficient of y), an integrating factor is
(∫
µ = exp
)
1 · dt = e t (4.49)
Multiplying the differential equation through by the integrating factor gives
d (
e t y ) = e t (y ′ + y) = e t cos t (4.50)
dt
Multiplying by dt and integrating,
∫ d (
e t y ) ∫
dt =
dt
e t cos tdt (4.51)
The left hand side is an exact derivative; the right hand side we can find
from an integral table or WolframAlpha:
∫
ye t = e t cos tdt = 1 2 et (sin t + cos t) + C (4.52)
The initial condition tells us that y = 1 when t = 0,
(1)e 0 = 1 2 e0 (sin(0) + cos(0)) + C (4.53)
1 = 1 2 + C (4.54)
C = 1 2
(4.55)
Substituting C = 1/2 into equation 4.52 gives
ye t = 1 2 et (sin t + cos t) + 1 2
(4.56)
We can then solve for y by dividing by e t ,
y = 1 2 (sin t + cos t) + 1 2 e−t (4.57)
which is the unique solution to the initial value problem.
Example 4.4. Solve the initial value problem
}
y ′ − 2y = e 7t
y(0) = 1
(4.58)