Solve at (x,y) = (0,1) $$ (x+1)dx+e^ydy=0 $$ $$ (x+1)dx=-e^ydy $$ $$ \int x*dx + \int 1*dx=-\int e^y*dy $$ $$ \frac{x^2}{2} + x + C = -e^y + K$$
For (x,y) = (0,1) $$ 1 = -e^1 + K = \frac{0^2}{2}+0+C$$ Where $K = 3.7183$ and $ C = 1$ $$ $$ My answer is $\frac{x^2}{2} + x + 3.7183 = -e^y + 1 $. I have 0 confidence that I did the right things.
There's only 1 example in my book that resembles this problem and it's not really helpful... So I appreciate the help if anyone could help me solve it. Thank you.