Given the following rst-order IVP: (e x + y) dx + (2 + x + yey ) dy = 0 y (0) = 1 (a) Show that this equation is exact. (b) Solve the exact equation for f(x, y) = C. (c) Apply the given initial conditions to nd the value of C that satises the IVP. (d) Check your answer by showing that the given DE above is the dierential of the equation f (x, y) = C (where C is whatever value you got from part (c)), and that the initial condition is satised.

let M=ex +y and N=2 +x +yey

(a) σM/σy =1 and σN/σx = 1

since σM/σy = σN/σx , it is an exact equation

(b) ∫M dx + ∫terms of N not containing x

∫(ex + y) dx +∫yey + 2 dy

xy + ex +yey -ey +2y=C

(c) using y(0)=1

C=3

(d) from the differential equation given

by dividing through by dx

dy/dx = (-y-ex) /(2+x+yey)

from the solution

(xy + ex +yey -ey =2y)=(3)

x + y + ex + yey + ey - ey + 2 = 0

= (-y-ex) /(2+x+yey)

Step-by-step explanation:

1. integrate with respect to x keeping y constant

2. integrate terms without x in N

3. Result of 1 + result 2= C

4. insert the condition given into 3

5.  compare the solution of 4 to the differential equation