Solve the differential equation $\frac{dy}{dx}=\frac{x+y-2}{x-y+4}$ with the initial condition $y(1)=1$.

Solution

1. Find the intersection of $x+y-2=0$ and $x-y+4=0$, which is $(-1,3)$.
2. Substitute $x=X-1$ and $y=Y+3$ to get $\frac{dY}{dX}=\frac{X+Y}{X-Y}$.
3. Use $Y=vX$ to obtain $$\begin{array}{rl}\frac{dv}{dx}X& =\frac{1+v}{1-v}-v\\ & =\frac{1+v^2}{1-v}\end{array}$$.
4. Integrate $\int \frac{1-v}{1+v^2}dv=\int \frac{dX}{X}$ to get ${\tan }^{-1}v-\frac{1}{2}\ln (1+v^2)=\ln X+C$.
5. Substitute back $v=\frac{y-3}{x+1}$ and use $(1,1)$ to find $C$, hence the answer is (B).