Lecture Notes in Differential Equations - Bruce E. Shapiro

86 LESSON 10. INTEGRATING FACTORS
Equating the right hand sides of (10.88) and (10.90) gives
t 3 + t 2 y + h ′ (y) = t 3 + t 2 y (10.91)
Therefore h ′ (y) = 0 and h(y) is a constant. From (10.89), general solution
of the differential equation, which is φ = C, is given by
t 3 y + 1 2 t2 y 2 = C. (10.92)
Example 10.3. Solve the initial value problem
}
(2y 3 + 2)dt + 3ty 2 dy = 0
y(1) = 4
(10.93)
This has the form M(t, y)dt + N(t, y)dy with
Since
To find an integrating factor, we start with
P (t, y) = M y − N t
N
M(t, y) = 2y 3 + 2 (10.94)
N(t, y) = 3ty 2 (10.95)
M y = 6y 2 (10.96)
N t = 3y 2 (10.97)
= 6y2 − 3y 2
3ty 2
= 3y2
3ty 2 = 1 t
(10.98)
This is only a function of t and so an integrating factor is
(∫ ) 1
µ = exp
t dt = exp (ln t) = t (10.99)
Multiplying (10.93) by µ(t) = t,
which has
(2ty 3 + 2t)dt + 3t 2 y 2 dy = 0 (10.100)
M(t, y) = 2ty 3 + 2t (10.101)
N(t, y) = 3t 2 y 2 (10.102)
Since
M y (t, y) = 6ty 2 = N t (t, y) (10.103)