Lecture Notes in Differential Equations - Bruce E. Shapiro

75
This has the form Mdt + Ndy = 0 with
M(t, y) = 2ye 2t + 2t cos y (9.105)
N(t, y) = e 2t − t 2 sin y (9.106)
Since
M y = 2e 2t − 2 sin y = N t (9.107)
we conclude that (9.104) is exact. Therefor the solution of the differential
is equation φ(t, y) = C for some constant C, where
Integrating (9.108) over t
∂φ
∂t = M(t, y) = 2ye2t + 2t cos y (9.108)
∂φ
∂y = N(t, y) = e2t − t 2 sin y (9.109)
∫
φ(t, y) =
(2ye 2t + 2t cos y)dt + h(y) = ye 2t + t 2 cos y + h(y) (9.110)
Taking the partial derivative with respect to y gives
∂φ
∂y = e2t − t 2 sin y + h ′ (y) (9.111)
Comparison of equations (9.111) and (9.109) gives h ′ (y) = 0; hence h(y) is
constant. Therefore
φ(t, y) = ye 2t + t 2 cos y (9.112)
and the general solution is
From the initial condition y(0) = 1,
ye 2t + t 2 cos y = C (9.113)
(1)(e 0 ) + (0 2 ) cos 1 = C =⇒ C = 1 (9.114)
and therefore the solution of the initial value problem is
ye 2t + t 2 cos y = 1. (9.115)