Lecture Notes in Differential Equations - Bruce E. Shapiro

30 LESSON 4. LINEAR EQUATIONS
Example 4.2. Solve ty ′ − y = t 2 e −t .
We first need to put this into standard form y ′ + p(t)y = q(t). If we divide
the differential equation on both sides by t then
y ′ − 1 t y = te−t (4.39)
Hence
p(t) = − 1 t
(4.40)
An integrating factor is
(∫
µ(t) = exp − 1 ) (
t dt = exp(− ln t) = exp ln 1 )
= 1 t t
(4.41)
Multiplying the differential equation (in standard form) through by µ(t),
1
(y − 1 ) ( ) 1
t t y = (te −t ) = e −t (4.42)
t
The left hand side is the exact derivative of µy:
Hence
d
dt µy = d ( y
)
= ty′ − y
dt t t 2 = 1 (
y ′ − y )
t t
(4.43)
d
( y
)
= e −t (4.44)
dt t
Multiplying by dt and integrating,
∫ d
( y
) ∫
dt =
dt t
e −t dt (4.45)
Since the left hand side is an exact derivative,
∫
y
t = e −t dt + C = −e −t + C (4.46)
Solving for y,
y = −te −t + Ct (4.47)
Example 4.3. Solve the initial value problem
y ′ + y = cos t, y(0) = 1 (4.48)