Question

Given that $\left( {{x}^{2}}+x+1 \right)dy+\left( {{y}^{2}}+y+1 \right)dx=0$. The constant value of the equation is C. What is C equal to?
\[\begin{align}
  & A.1 \\
 & B.-1 \\
 & C.2 \\
 & D.\text{None of the above} \\
\end{align}\]

Answer 1

Hint: At first, separate the terms according to the variables, then use the following identity,
\[\int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\]
Where ‘a’ is any constant. Then, simplify it further and finally compare it with the given general equation to find the desired answer.

Complete step-by-step solution:
In the above question we are given a differential equation as
\[\left( {{x}^{2}}+x+1 \right)dy+\left( {{y}^{2}}+y+1 \right)dx=0\]
And its constant value of the main equation is C and we have to find its value.
So, we are given the following equation which is,
\[\left( {{x}^{2}}+x+1 \right)dy+\left( {{y}^{2}}+y+1 \right)dx=0\]
Which can also be written as,
\[\left( {{x}^{2}}+x+1 \right)dy=-\left( {{y}^{2}}+y+1 \right)dx\]
Which can be further written as,
\[\int{\dfrac{dy}{\left( {{y}^{2}}+y+1 \right)}}=\int{-\dfrac{dx}{\left( {{x}^{2}}+x+1 \right)}}\]
We can write the given expression ${{y}^{2}}+y+1$ as ${{\left( y+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}\Rightarrow {{\left( y+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$ and ${{x}^{2}}+x+1$ as ${{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}\Rightarrow {{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$
So we can write the equation as,
\[\int{\dfrac{dy}{{{\left( y+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}}=\int{-\dfrac{dx}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}}\]
Now we all use a formula which is,
\[\int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\]
Where, ‘a’ is constant. On applying this, we get:
\[\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{y+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \right)=\dfrac{-2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \right)+C\]
Let’s take C as $\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}g$ which is a constant for ease of simplifying, so we get:
\[\begin{align}
  & \dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2y+1}{\sqrt{3}} \right)=\dfrac{-2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right)+\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}g \\
 & \Rightarrow \dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right)+\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2y+1}{\sqrt{3}} \right)=\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}g \\
 & \Rightarrow {{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right)+\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2y+1}{\sqrt{3}} \right)=\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}g \\
\end{align}\]
Now using identity,
\[{{\tan }^{-1}}C+{{\tan }^{-1}}D=\tan -1\left( \dfrac{C+D}{1-CD} \right)\]
So we get,
\[{{\tan }^{-1}}\left( \dfrac{\dfrac{2x+1}{\sqrt{3}}+\dfrac{2y+1}{\sqrt{3}}}{1-\dfrac{\left( 2x+1 \right)\left( 2y+1 \right)}{3}} \right)={{\tan }^{-1}}g\]
So we can write it as,
\[\begin{align}
  & \dfrac{\dfrac{2\left( x+y+1 \right)}{\sqrt{3}}}{\dfrac{3-4xy-1-2x-2y}{3}}=g \\
 & \Rightarrow \dfrac{2}{\sqrt{3}}\left( x+y+1 \right)=\dfrac{g}{3}\left( 2-2x-2y-4xy \right) \\
 & \Rightarrow 2\left( x+y+1 \right)=\dfrac{g}{\sqrt{3}}\left( 2-2x-2y-4xy \right) \\
 & \Rightarrow \left( x+y+1 \right)=\dfrac{g}{2\sqrt{3}}\left( 2-2x-2y-4xy \right) \\
 & \Rightarrow \left( x+y+1 \right)=\dfrac{g}{\sqrt{3}}\left( 1-x-y-2xy \right) \\
\end{align}\]
We were given the general solution as,
\[x+y+1=A\left( 1+Bx+Cy+Dxy \right)\]
Hence, on comparing we can say that C is -1.
Thus, the correct option is B.

Note: Students while integrating should be careful about the coefficient or constant in the expression. They sometimes leave out the constant of integration which can make the answer wrong. So, this is the main point of which they should take care off.