Lecture Notes in Differential Equations - Bruce E. Shapiro

87
the revised equation (10.100) is exact and therefore the solution is φ(t, y) =
C where
∂φ
∂t = M(t, y) = 2ty3 + 2t (10.104)
∂φ
∂y = N(t, y) = 3t2 y 2 (10.105)
Integrating (10.104) with over t,
∫ ∂φ
φ(t, y) = dt + h(y) (10.106)
∂t
∫
= (2ty 3 + 2t)dt + h(y) (10.107)
Differentiating with respect to y,
= t 2 y 3 + t 2 + h(y) (10.108)
∂φ(t, y)
∂y
= 3t 2 y + h ′ (y) (10.109)
Equating the right hand sides of (10.109) and (10.105) gives
3t 2 y 2 + h ′ (y) = 3t 2 y 2 (10.110)
Hence h ′ (y) = 0 and h(y) = C for some constant C. From (10.108) the
general solution is
t 2 (y 3 + 1) = C (10.111)
Applying the initial condition y(1) = 4,
(1) 2 (4 3 + 1) = C =⇒ C = 65 (10.112)
Therefore the solution of the initial value problem (10.93) is
t 2 (y 3 + 1) = 65 (10.113)
Example 10.4. Solve the initial value problem
}
ydt + (2t − ye y )dy = 0
y(0) = 1
(10.114)
Equation (10.114) has the form M(t, y)dt + N(t, y)dy = 0 with M(t, y) = y
and N(t, y) = 2t − ye y . Since M y = 1 ≠ 2 = N t the differential equation is
not exact. We first try case 1:
P (1) (t, y) = M y(t, y) − N t (t, y)
N(t, y)
=
1
ye y − 2t
(10.115)