Lecture Notes in Differential Equations - Bruce E. Shapiro

88 LESSON 10. INTEGRATING FACTORS
This depends on both t and y; case 1 requires that P (1) (t, y) only depend
on t. Hence case 1 fails. Next we try case 2
P (2) (t, y) = N t(t, y) − M y (t, y)
M(t, y)
= 1 y
(10.116)
Since P (2) is purely a function of y, the conditions for case (2) are satisfied.
Hence an integrating factor is
(∫ )
µ(y) = exp (1/y)dy = exp (ln y) = y (10.117)
Multiplying equation the differential equation through µ(y) = y gives
This has
y 2 dt + (2ty − y 2 e y )dy = 0 (10.118)
M(t, y) = y 2 (10.119)
N(t, y) = 2ty − y 2 e y (10.120)
Since M y = 2y = N t , equation (10.118) is exact. Thus we know that the
solution is φ(t, y) = C where
∂φ
∂t =M(t, y) = y2 (10.121)
∂φ
∂y =N(t, y) = 2ty − y2 e y (10.122)
Integrating (10.121) over t,
∫ ∫
∂φ
φ(t, y) =
∂t dt + h(y) = y 2 dt + h(y) = y 2 t + h(y) (10.123)
Differentiating with respect to y,
∂φ
∂y = 2ty + h′ (y) (10.124)
Equating the right hand sides of (10.124) and (10.122),
2yt + h ′ (y) = 2ty − y 2 e y (10.125)
h ′ (y) = −y 2 e y
∫
(10.126)
h(y) = h ′ (y)dy (10.127)
= −
∫
y 2 e y dy (10.128)
= −e y (2 − 2y + y 2 ) (10.129)