Question
The solutions of $$\left( {x + y + 1} \right)dy = dx$$ are :
A.
$$x + y + 2 = C{e^y}$$
B.
$$x + y + 4 = C\,\log \,y$$
C.
$$\log \left( {x + y + 2} \right) = Cy$$
D.
$$\log \left( {x + y + 2} \right) = C - y$$
Answer :
$$x + y + 2 = C{e^y}$$
Solution :
Putting $$x + y + 1 = u,$$ we have $$du = dx + dy$$ and the given equation reduces to $$u\left( {du - dx} \right) = dx$$
$$\eqalign{
& \Rightarrow \frac{{u\,du}}{{u + 1}} = dx \cr
& \Rightarrow u - \log \left( {u + 1} \right) = x \cr
& \Rightarrow \log \left( {x + y + 2} \right) = y + {\text{constant}} \cr
& \Rightarrow x + y + 2 = C{e^y} \cr} $$