The general solution of the differential equa | vedclass

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(A) Given differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x-y-1}$.Let $x = X+h$ and $y = Y+k$,then $\frac{dy}{dx} = \frac{dY}{dX}$.Substituting t

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$	an^{-1}\left(\frac{y+1}{x}ight) - \frac{1}{2} \log(x^2+y^2+2y+1) = c$

Vedclass · Answer

(A) Given differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x-y-1}$.Let $x = X+h$ and $y = Y+k$,then $\frac{dy}{dx} = \frac{dY}{dX}$.Substituting these,we get $\frac{dY}{dX} = \frac{X+Y+h+k+1}{X-Y+h-k-1}$.For the equation to be homogeneous,we set $h+k+1 = 0$ and $h-k-1 = 0$.Solving these equations,we find $h = 0$ and $k = -1$.Thus,the equation becomes $\frac{dY}{dX} = \frac{X+Y}{X-Y}$.Rearranging,we get $(X-Y) dY = (X+Y) dX$,which implies $X dY - Y dX = X dX + Y dY$.Dividing by $X^2+Y^2$,we have $\frac{X dY - Y dX}{X^2+Y^2} = \frac{X dX + Y dY}{X^2+Y^2}$.Integrating both sides,$\int d\left(	an^{-1}\left(\frac{Y}{X}ight)ight) = \frac{1}{2} \int d(\log(X^2+Y^2))$.This gives $	an^{-1}\left(\frac{Y}{X}ight) = \frac{1}{2} \log(X^2+Y^2) + C$.Substituting $X = x$ and $Y = y+1$,we get $	an^{-1}\left(\frac{y+1}{x}ight) = \frac{1}{2} \log(x^2+(y+1)^2) + C$.Therefore,$	an^{-1}\left(\frac{y+1}{x}ight) - \frac{1}{2} \log(x^2+y^2+2y+1) = C$.

The general solution of the differential equation $(x-y-1) dy = (x+y+1) dx$ is

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