# Solve $(x+1)dx+e^ydy=0$ at (x,y)=(0,1)

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 only need one constant of integration, because they can just combine into a single one. Jun 12, 2016 at 23:28
• The integration looks good, but you are wrong at the last step. You shall plug x=0 and y=1 into the expression involved with C and K, and then you get C-K =-e. From here you can continue to simplify the expression. Jun 12, 2016 at 23:33

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: $$y=\ln\left(-\frac{x^2}{2}-x+e\right)$$
• @Majid It doesn't matter whether or not we use one or two integration constants. In fact, we could say $x^2+x+C_1+C_2=-e^y+K_1+K_2+K_3$ and still come up with the correct answer. What matters is that we find that all of the constants on the left side subtracted by all of the constants on the right side is $-e$. Personally, I think this is easiest if we only use one integration constant, but I went off of the work of the original questioner, who used two integration constants. Jun 13, 2016 at 1:03
• @Majid OK, I added an explanation that we only need to use one integration constant called it $D$. Jun 13, 2016 at 1:27