Differential Equations

To solve the given differential equation, we start by analyzing the equation:$(x-y-1) dy=(x+y+1) dx$This can be rewritten in differential form as:$\frac{dy}{dx}=\frac{x+y+1}{x-y-1}$This expression can also be represented as:$\frac{dy}{dx}=\frac{x+(y+1)}{x-(y+1)}$Introduce a substitution where $y+1=z$. Thus, we have:$\frac{dy}{dx}=\frac{dz}{dx}$After substitution, the equation becomes:$\frac{dz}{dx}=\frac{x+z}{x-z}$Next, we use another substitution $z=v x$. Therefore, we differentiate $z$ with respect to $x$ :$\frac{dz}{dx}=v+x\frac{dv}{dx}$Substituting this back into our equation results in:$v+x\frac{dv}{dx}=\frac{1+v}{1-v}$Reorganize terms and separate variables:$x\frac{dv}{dx}=\frac{1+v-v+v^2}{1-v}$This simplifies to:$\left(\frac{1-v}{1+v^2}\right) dv=\frac{dx}{x}$Integrating both sides, we have:$\int \left( \frac{1}{1+v^2} - \frac{v}{1+v^2} \right) dv = \int \frac{dx}{x}$The integrals yield: $\tan^{-1} v - \frac{1}{2} \log \left| 1+v^2 \right| = \log x + C$Substitute back $v=\frac{z}{x}$ :$\tan^{-1} \frac{z}{x} - \frac{1}{2} \log \left| \frac{z^2}{x^2} \right| = \log x + C$Simplifying using $z=y+1$ :$\tan^{-1} \left( \frac{y+1}{x} \right) - \frac{1}{2} \log \left| \frac{x^2+(y+1)^2}{x^2} \right| - \log x = C$Finally, this simplifies to:$\tan^{-1} \left( \frac{y+1}{x} \right) - \frac{1}{2} \log \left| x^2+y^2+2y+1 \right| = C$