Q.Show that the general solution of the differential equation $\dfrac{dy}{dx}=\dfrac{y^2+y+1}{x^2+x+1}$ is given by $(x+y+1)=c(1-x-y-2xy)$.

$\dfrac{dy}{dx}=\dfrac{y^2+y+1}{x^2+x+1}$ separates as $\dfrac{dy}{y^2+y+1}=\dfrac{dx}{x^2+x+1}$. Both sides are of the standard form $\int\dfrac{dt}{t^2+t+1}=\dfrac{2}{\sqrt3}\tan^{-1}\!\left(\dfrac{2t+1}{\sqrt3}\right)+c$, so integrating gives $\tan^{-1}\!\left(\dfrac{2y+1}{\sqrt3}\right)-\tan^{-1}\!\left(\dfrac{2x+1}{\sqrt3}\right)=k$ for an arbitrary constant $k$. Taking the tangent of both sides and simplifying algebraically gives the rational closed form $\dfrac{y-x}{2+x+y+2xy}=c$ (an arbitrary constant $c=\tan k/\sqrt3$), which is directly verified by implicit differentiation to satisfy the original equation exactly. (Cross-checking the printed targe …

Separate the variables (both sides reduce to ∫dt/(t²+t+1), a standard arctan integral), integrate, and simplify the arctan-difference into a rational closed form using the tangent-subtraction id …

Assuming the printed target formula must be correct without independently verifying it by implicit differentiation — always re-derive and check, s …