Problem 12
Solve the differential equation x+xy+y'(y+xy)=0
Solve the differential equation $x+xy+y'(y+xy)=0$.
Explanation of Method
Factor the equation to separate the $x$- and $y$-dependence, rewrite each side using $\dfrac{u}{1+u}=1-\dfrac{1}{1+u}$, and integrate.
Step 1: Factor and separate
Factoring, $x(1+y)+yy'(1+x)=0$, so (for $1+x,1+y\neq0$)
\[\frac{x}{1+x}\,dx=-\frac{y}{1+y}\,dy.\]
Step 2: Integrate
Using $\dfrac{x}{1+x}=1-\dfrac{1}{1+x}$ and similarly for $y$,
\[x-\ln|1+x|=-y+\ln|1+y|+C \quad\Longrightarrow\quad x+y-\ln|(1+x)(1+y)|=C.\]
Exponentiating,
\[\boxed{\,(1+x)(1+y)=Ce^{x+y}.}\]
Step 3: Domain remarks
The derivation requires $x\neq-1$ and $y\neq-1$, so these lines are excluded from (and treated separately from) the solution family.
Result
Final answer:
\[\boxed{\,(1+x)(1+y)=Ce^{x+y},\qquad x\neq-1,\ y\neq-1.}\]