Question

solve the given differential equation by using an appropriate substitution. the de is homogeneous.
\\(\frac{dy}{dx}=\frac{y - x}{y + x}\\)

Explanation:

Step1: Use the substitution $y = vx$

If $y = vx$, then $\frac{dy}{dx}=v + x\frac{dv}{dx}$. The given differential - equation $\frac{dy}{dx}=\frac{y - x}{y + x}$ becomes $v + x\frac{dv}{dx}=\frac{vx - x}{vx + x}$.

Step2: Simplify the right - hand side

Simplify $\frac{vx - x}{vx + x}$ by factoring out $x$ from the numerator and denominator: $\frac{vx - x}{vx + x}=\frac{x(v - 1)}{x(v + 1)}=\frac{v - 1}{v + 1}$. So, $v + x\frac{dv}{dx}=\frac{v - 1}{v + 1}$.

Step3: Rearrange the equation

Rearrange the equation $v + x\frac{dv}{dx}=\frac{v - 1}{v + 1}$ to get $x\frac{dv}{dx}=\frac{v - 1}{v + 1}-v$. Combine the terms on the right - hand side: $x\frac{dv}{dx}=\frac{v - 1-v(v + 1)}{v + 1}=\frac{v - 1 - v^{2}-v}{v + 1}=\frac{-v^{2}-1}{v + 1}$.

Step4: Separate the variables

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

Step5: Integrate both sides

Integrate $\int\frac{v + 1}{v^{2}+1}dv=-\int\frac{1}{x}dx$. Split $\int\frac{v + 1}{v^{2}+1}dv$ into $\int\frac{v}{v^{2}+1}dv+\int\frac{1}{v^{2}+1}dv$.
For $\int\frac{v}{v^{2}+1}dv$, 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, $\frac{1}{2}\ln(v^{2}+1)+\arctan(v)=-\ln|x|+C$.

Step6: Substitute back $v=\frac{y}{x}$

Substitute $v = \frac{y}{x}$: $\frac{1}{2}\ln(\frac{y^{2}}{x^{2}}+1)+\arctan(\frac{y}{x})=-\ln|x|+C$.
We can rewrite $\frac{1}{2}\ln(\frac{y^{2}+x^{2}}{x^{2}})+\arctan(\frac{y}{x})=-\ln|x|+C$.
Using the property of logarithms $\frac{1}{2}(\ln(y^{2}+x^{2})-\ln(x^{2}))+\arctan(\frac{y}{x})=-\ln|x|+C$.
$\frac{1}{2}\ln(x^{2}+y^{2})-\ln|x|+\arctan(\frac{y}{x})=-\ln|x|+C$.

Answer:

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