Lecture Notes in Differential Equations - Bruce E. Shapiro

68 LESSON 9. EXACT EQUATIONS
and
Since
the equation is exact, and therefore the solution is
where
and
∂N
∂t = ∂ (2ty) = 2y (9.28)
∂t
M y = N t (9.29)
φ = C (9.30)
∂φ
∂t = M = 2t + y2 (9.31)
∂φ
∂y
= N = 2ty (9.32)
To solve the original ODE (9.25) we need to solve the two partial differential
equations (9.31) and (9.32). Fortunately we can illustrate a procedure to
do this that does not require any knowledge of the methods of partial
differential equations, and this method will always work.
Method 1. Start with equation (9.31)
∂φ
∂t = 2t + y2 (9.33)
Integrate both sides over t (because the derivative is with respect to t),
treating y as a constant in the integration. If we do this, then the constant
of integration may depend on y:
∫ ∫ ∂φ
∂t dt = (2t + y 2 )dt (9.34)
φ = t 2 + y 2 t + g(y) (9.35)
where g(y) is the constant of integration that depends on y. Now substitute
(9.35) into (9.32)
∂
∂y (t2 + y 2 t + g(y)) = 2ty (9.36)
Evaluating the partial derivatives,
2yt + ∂g(y)
∂y
= 2ty (9.37)
Since g(y) only depends on y, then the partial derivative is the same as
g ′ (y):
2yt + dg = 2ty (9.38)
dy