Lecture Notes in Differential Equations - Bruce E. Shapiro

73
Differentiating equations (9.81) and (9.82) gives
∂M(t, y)
∂y
= cos t + 2te y =
∂N(t, y)
∂t
and consequently equation (9.79) is exact. Hence the solution is
(9.83)
φ(t, y) = C (9.84)
where
∂φ
∂t = M(t, y) = y cos t + 2tey (9.85)
∂φ
∂y = N(t, y) = sin t + t2 e y + 2 (9.86)
Integrating equation (9.85) over t
∫ ∫
∂φ(t, y)
φ(t, y) = dt = (y cos t + 2te y ) dt = y sin t+t 2 e y +h(y) (9.87)
∂t
where h is an unknown function of y.
Differentiating (9.87) respect to y and setting the result equal to (9.86)
gives
Thus
∂φ
∂y = sin t + t2 e y + h ′ (y) = sin t + t 2 e y + 2 (9.88)
h ′ (y) = 2 =⇒ h(y) = 2y (9.89)
Using this back in equation (9.87)
φ(t, y) = y sin t + t 2 e y + 2y (9.90)
Hence the required family of solutiosn is
y sin t + t 2 e y + 2y = C (9.91)
for any value of the constant C.
Example 9.5. Solve the differential equation
y ′ =
at − by
bt + cy
where a, b, c, and d are arbitrary constants.
(9.92)