When trying to find the original function for this one, we have the quantity of X plus YE to the YDYDX equalling one. We could distribute the DY to the DYDX to the two terms, but we need to get it in a form that we're familiar with. So we really want DYDX to have A1 coefficient eventually minus the Y variable times some coefficient. So when we look at this, we can see that's not going to be easily done. So instead of doing it in dy DX, let's try to do it in DX dy. So the first thing we're going to do is we're going to multiply each side by DX dy. Then we can see I can subtract the single X over, and now I have DX dy with A1 coefficient minus something times a plain old X equaling YE to the Y. So now we're going to figure out what our row is. So this time it's going to be row of Y instead of row of X equaling E to the integral -1 the coefficient on the X term DY. So when we evaluate that, we get E to the negative Y, then we're going to take that term and we're going to multiply it all the way through. So now we have E to the negative YDYD X -, e to the negative yx equaling Y because the E to the Y and the E to the negative Y are going to cancel, giving us one. Now the two terms on the left hand side are really going to be the product rule for E to the negative yx, taking the derivative in terms of Y. So ddy of E to the negative yx is going to equal plane Y. We're going to take and we're going to evaluate the integral of each side. Remember fundamental theorem of calculus as the integral of a derivative are going to cancel. So we get E to the negative yx equaling the integral of YDY. Add 1 to the exponent, divide by the new exponent on the right hand side and make sure to add the constant. So then to get X by itself, we're going to multiply in the entire equation through by east to the Y or divide it by east to the negative Y. For final answer of X equal 1 half y ^2 east to the Y + C, E to the Y.