Lecture Notes in Differential Equations - Bruce E. Shapiro

84 LESSON 10. INTEGRATING FACTORS
By the chain rule,
∂µ
∂t = dµ ∂z
dz ∂t = µ′
y
(10.67)
∂µ
∂y = dµ ∂z
dz ∂y = t
−µ′ y 2 (10.68)
where µ ′ (z) = du/dz. Substituting this into equation (10.38) gives
( )
( −µ(z)t
µ ′ )
(z)
µ(z)M y (t, y) +
y 2 M(t, y) = µ(z)N t (t, y) + N(t, y)
y
(10.69)
Rearranging
( µ ′ ) ( )
(z)
µ(z)t
µ(z)(M y (t, y) − N t (t, y)) = N(t, y) +
y
y 2 M(t, y) (10.70)
= µ′ (z)
(yN(t, y) + tM(t, y)) (10.71)
y2 µ ′ (z)
µ(z) = y2 (M y (t, y) − N t (t, y))
= P (5) (z) (10.72)
yN(t, y) + tM(t, y)
Multiplying by dz, integrating, and exponentiating gives
(∫ )
µ(z) = exp P (5) (z)dz
(10.73)
as required by equation (10.35). (Case 5)
This completes the proof for all five case.
Example 10.2. . Solve the differential equation
(3ty + y 2 ) + (t 2 + ty)y ′ = 0 (10.74)
by finding an integrating factor that makes it exact.
This equation has the form Mdt + Ndy where
M(t, y) = 3ty + y 2 (10.75)
N(t, y) = t 2 + ty (10.76)
First, check to see if the equation is already exact.
M y = 3t + 2y (10.77)
N t = 2t + y (10.78)