Lecture Notes in Differential Equations - Bruce E. Shapiro

76 LESSON 9. EXACT EQUATIONS
Example 9.7. Solve the initial value problem
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(y/t + cos t)dt + (e y + ln t)dy = 0
y(π) = 0
(9.116)
The ODE has the form Mdt + Ndy = 0 with
Checking the partial derivatives,
M(t, y) = y/t + cos t (9.117)
N(t, y) = e y + ln t (9.118)
M y = 1 t = N t (9.119)
hence the differential equation is exact. Thus the solution is φ(t, y) = C
where
Integrating the first of these equations
φ(t, y) =
∂φ
∂t = M(t, y) = y + cos t
t
(9.120)
∂φ
∂y = N(t, y) = ey + ln t (9.121)
∫ (y
t + cos t )
dt + h(y) = y ln t + sin t + h(y) (9.122)
Differentiating with respect to y and setting the result equal to (9.121),
From (9.122),
∂φ
∂y = ∂ ∂y (y ln t + sin t + h(y)) = N(t, y) = ey + ln t (9.123)
ln t + h ′ (y) = e y + ln t (9.124)
h ′ (y) = e y =⇒ h(y) = e y (9.125)
φ(t, y) =) = y ln t + sin t + e y (9.126)
The general solution of the differential equation is φ(t, y) = C, i.e.,
The initial condition is y(π) = 0; hence
Thus
y ln t + sin t + e y = C (9.127)
(0) ln π + sin π + e 0 = C =⇒ C = 1 (9.128)
y ln t + sin t + e y = 1. (9.129)