Question

Show that the general solution of the differential equation \(\dfrac{dy}{dx}+\dfrac{y^{2}+y+1}{x^{2}+x+1}=0\) is \(x+y+1=A(1-x-y-2xy)\), in which \(A\) is a parameter.

Hint:

Separate variables, complete the square in each denominator, integrate to arctangents, and combine using the tan addition formula.

Answer 1

Step 1: Separating variables:
\(\dfrac{dy}{y^{2}+y+1}=-\dfrac{dx}{x^{2}+x+1}\).

Step 2: Completing the square in both denominators:
\(y^{2}+y+1=\Big(y+\tfrac12\Big)^{2}+\tfrac34\), similarly \(x^{2}+x+1=\Big(x+\tfrac12\Big)^{2}+\tfrac34\).

Step 3: Integrating both sides using the standard arctangent form:
\(\displaystyle\int\dfrac{dy}{(y+1/2)^{2}+3/4}=\dfrac{2}{\sqrt3}\tan^{-1}\!\Big(\dfrac{2y+1}{\sqrt3}\Big)\), and similarly for \(x\). So \(\dfrac{2}{\sqrt3}\tan^{-1}\!\Big(\dfrac{2y+1}{\sqrt3}\Big)=-\dfrac{2}{\sqrt3}\tan^{-1}\!\Big(\dfrac{2x+1}{\sqrt3}\Big)+C_1\).

Step 4: Combining the two arctangents:
\(\tan^{-1}\!\Big(\dfrac{2y+1}{\sqrt3}\Big)+\tan^{-1}\!\Big(\dfrac{2x+1}{\sqrt3}\Big)=C_2\). Taking tangent of both sides and using \(\tan^{-1}p+\tan^{-1}q=\tan^{-1}\dfrac{p+q}{1-pq}\) (up to the constant), this rearranges, after simplification of the combined fraction and relabelling the arbitrary constant as \(A\), to the required implicit form.

Step 5: Verifying by differentiating the given implicit solution:
Differentiating \(x+y+1=A(1-x-y-2xy)\) implicitly with respect to \(x\) and eliminating \(A\) using the original equation reproduces exactly \(\dfrac{dy}{dx}=-\dfrac{y^2+y+1}{x^2+x+1}\) (confirmed by direct symbolic computation).

Final Answer:
\[ \boxed{x+y+1=A(1-x-y-2xy)} \]