Lecture Notes in Differential Equations - Bruce E. Shapiro

37
∫ t
t 0
Evaluating the integral,
∫
d
t
ds (µ(s)y(s))ds =
µ(t)y(t) − µ(t 0 )y(t 0 ) =
t 0
µ(s)q(s)ds (4.94)
∫ t
t 0
µ(s)q(s)ds (4.95)
Solving for y(t) gives a general formula for the solution of a linear initial
value problem:
y(t) = 1 [
∫ t
]
µ(t 0 )y(t 0 ) + µ(s)q(s)ds
µ(t)
t 0
Example 4.8. Use equation 4.96 to solve
t dy
dt + 2y = 4 ⎫
t sin t ⎬
⎭
y(π) = 0
(4.96)
(4.97)
Rewriting in standard form for a linear differential equation,
y ′ + 2 t y = 4 sin t (4.98)
t2 Hence p(t) = 2/t and q(t) = (4/t) 2 sin t. An integrating factor is
(∫ ) 2
µ(t) = exp
t dt = exp (2 ln t) = t 2 (4.99)
Thus equation 4.96 gives us
y(t) = 1 t 2 [(π 2 )(0) +
∫ t
π
(s 2 )
( ) ]
4
s 2 sin s ds
(4.100)
= 4 ∫ t
t 2 sin s ds (4.101)
π
4(cos t − cos π)
= −
t 2 (4.102)
4(cos t + 1)
= −
t 2 (4.103)