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 substitution $y = vx$

Since $y = vx$, 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 $\frac{vx - x}{vx + x}=\frac{x(v - 1)}{x(v + 1)}=\frac{v - 1}{v + 1}$. So, $v + x\frac{dv}{dx}=\frac{v - 1}{v + 1}$.

Step3: Rearrange the equation

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

Step4: Separate the variables

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

Step5: Integrate both sides

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

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

Answer:

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