Anyone help out with this Diff EQ. question??? 10 points for any answer that has numbers in it!!!?

Question

(e^x+y) dx + (2+x+ye^y) = 0, y(0) = 1

Answer 1

I think you are missing a dy term somewhere, or is your dx actually a dy/dx ???? Can't help until all the information is correct ....

assuming you mean
(e^x + y)dx + ( 2 + x + ye^y)dy = 0
(e^x)dx + (y)dx + 2 dy+ xdy + ye^ydy = 0

grouping ydx + xdy ---> d(xy),
and integral of (ye^y)dy = (e^y) [y- 1 ] , then easy to integrate
[ e^x ]dx, and 2dy.
thus: e^x + 2y + xy + (e^y) [ y - 1 ] = C
using IC x = 0, y =1 ,
1 + 2 + 0 + e [ 1-1 ] = C , then C = 3
thus e^x + 2y + xy + (e^y) [ y - 1 ] = 3

check....differentiating with respect to x, and then multiplying the final equation thru by dx, gives the original problem I assumed above....

Answer 2

(e^x + y)dx + (2 + x + ye^y)dy = 0, y(0) = 1. This is exact because d(e^x + y)/dy = 1 = d(2 + x + ye^y)/dx. We are looking for a function G(x,y) such that dG/dx = e^x + y and dG/dy = 2 + x + ye^y.

From dG/dx = e^x + y we get G = e^x + xy + f(y). Then dG/dy = x + f'(y) = 2 + x + ye^y. Therefore, f'(y) = 2 + ye^y, so f(y) = 2y + ye^y - e^y. Thus, the function g(x,y) = e^x + xy + 2y +ye^y - e^y, and the solution of the diff. eq. is G(x,y) = C. Use the initial condition y = 1 when x = 0 to find C = 3.