Page 52 - Schaum's Outline of Differential Equations

CHAP. 5] EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 35

5.7. Determine whether the differential equation y 2 dt + (2yt + 1) dy = 0 is exact.
2
This is an equation for the unknown function y(t). In terms of the variables t and y, we have M(t, y) =y ,
I
N(t,y) = 2yt+ , and

so the differential equation is exact.

5.8. Solve the differential equation given in Problem 5.7.
This equation was shown to be exact, so the solution procedure given by Eqs. (5.4) through (5.6), with t
replacing x, is applicable. Here

Integrating both sides with respect to t, we have

Differentiating (1) with respect to y, we obtain

Hence,

where the right side of this last equation is the coefficient of dy in the original differential equation. It follows
that

2
h(y) =y + GI, and (_/) becomes g(t, y) = y t + y + Cj. The solution to the differential equation is given implicitly by
(5.6) as

We can solve for y explicitly with the quadratic formula, whence

5.9. Determine whether the differential equation

is exact.
This is an equation for the unknown function x(t). In terms of the variables t and x, we find

so the differential equation is exact.