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.
You are right in your integration, so we have that: $$\frac{x^2}{2}+x+C=-e^y+K$$ Notice that we have two constants on either side. If we subtract both sides by $K$: $$\frac{x^2}{2}+x+C-K=-e^y$$ Now, we have just one integration constant of $C-K$, so we don't need to work with two constants. We can just say $D=C-K$ and continue. This is a lot easier than using two integration constants, so you should always use one integration constant when you take the integral of both sides.
Now, when we substitute in $(x, y)=(0, 1)$, we get: $$\frac{0^2}{2}+0+D=-e^y$$ Notice the lack of a $1=$ in this equation. There is no reason for both of these expression to equal $1$ because they are not $y$. If we solve for $D$, we get $D=-e$
Back to our original equation: $$\frac{x^2}{2}+x+D=-e^y$$ Substitute $D=-e$: $$\frac{x^2}{2}+x-e=-e^y$$ Switch both sides and take the negative: $$e^y=-\frac{x^2}{2}-x+e$$ Take the natural log of both sides: $$y=\ln\left(-\frac{x^2}{2}-x+e\right)$$