solve the given differential equation by using an appropriate substitution. the de is homogeneous…

Explanation:

Step1: Rewrite the differential equation

Given $\frac{dy}{dx}=\frac{y - x}{y + x}$, we can rewrite it in the form of a homogeneous - differential equation. Let $y=vx$, then $\frac{dy}{dx}=v + x\frac{dv}{dx}$. Substituting $y = vx$ into the given differential equation, we get $v + x\frac{dv}{dx}=\frac{vx - x}{vx + x}=\frac{v - 1}{v + 1}$.

Step2: Rearrange the equation

$x\frac{dv}{dx}=\frac{v - 1}{v + 1}-v=\frac{v - 1 - v(v + 1)}{v + 1}=\frac{v - 1 - v^{2}-v}{v + 1}=\frac{-v^{2}-1}{v + 1}$.

Step3: Separate the variables

$\frac{v + 1}{v^{2}+1}dv=-\frac{1}{x}dx$.

Step4: Integrate both sides

Integrate $\int\frac{v + 1}{v^{2}+1}dv=-\int\frac{1}{x}dx$. We know that $\int\frac{v + 1}{v^{2}+1}dv=\int\frac{v}{v^{2}+1}dv+\int\frac{1}{v^{2}+1}dv$. Let $u = v^{2}+1$, then $du = 2vdv$ and $\int\frac{v}{v^{2}+1}dv=\frac{1}{2}\ln|v^{2}+1|$, and $\int\frac{1}{v^{2}+1}dv=\arctan(v)$. So, $\frac{1}{2}\ln(v^{2}+1)+\arctan(v)=-\ln|x|+C$.

Step5: Substitute back $v=\frac{y}{x}$

$\frac{1}{2}\ln(\frac{y^{2}}{x^{2}}+1)+\arctan(\frac{y}{x})=-\ln|x|+C$.

Answer:

$\frac{1}{2}\ln(\frac{y^{2}}{x^{2}} + 1)+\arctan(\frac{y}{x})=-\ln|x|+C$