The general solution of the differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ is

The general solution of the differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ is
  1. $x+y+1=c(1+x+y+2 x y)$
  2. $x+y+1=c(2+x+y+2 x y)$
  3. $x+y+1=c(1-x-y-2 x y$
  4. $x+y+2=c(2-x-y-2 x y)$

Solution

$\begin{aligned} & \frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0 \\ & \therefore \frac{d y}{d x}=-\left(\frac{y^2+y+1}{x^2+x-1}\right) \\ & \therefore \int \frac{d y}{y^2+y+1}=-\int \frac{d x}{x^2+x+1} \\ & \Rightarrow \int \frac{d y}{y^2+y+\frac{1}{4}+\frac{3}{4}}=-\int \frac{d x}{x^2+x+\frac{1}{4}+\frac{3}{4}} \\ & \int \frac{d y}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=-\int \frac{d x}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \\ & \Rightarrow \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} \tan ^{-1}\left[\frac{y+\frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)}\right]=\frac{-1}{\left(\frac{\sqrt{3}}{2}\right)^{-1} \tan ^{-1}}\left[\frac{x+\frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)}\right]+C_1\end{aligned}$ $\begin{aligned} & \therefore \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{y}+1}{\sqrt{3}}\right)=\frac{-2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)+\mathrm{C}_1 \\ & \therefore \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{\frac{(2 \mathrm{y}+1)}{\sqrt{3}}+\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)}{1-\left(\frac{2 \mathrm{y}+1}{\sqrt{3}}\right)\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)}\right]=\mathrm{C}_1 \\ & \Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{\frac{2(x+y+1)}{\sqrt{3}}}{\frac{3-(4 x y+2 x+2 y+1)}{3}}\right]=C_1 \\ & \therefore \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{2(x+y+1)}{\sqrt{3}} \times \frac{3}{2(1-2 x y-x-y)}\right]=C_1 \\ & \end{aligned}$ $\begin{aligned} & \therefore \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{\sqrt{3}(x+y+1)}{1-2 x y-x-y}\right]=C_1 \\ & \Rightarrow \tan ^{-1}\left[\frac{\sqrt{3}(x+y+1)}{1-x-y-2 x y}\right]=C_2 \\ & \ldots\left[\text { where } \mathrm{C}_2=\frac{\sqrt{3} \mathrm{C}_1}{2}\right] \\ & \therefore \frac{\sqrt{3}(x+y+1)}{1-x-y-2 x y}=\tan C_2 \Rightarrow \frac{x+y+1}{1-x-y-2 x y}=\frac{\tan C_2}{\sqrt{3}}=c \\ & \text {... (say) } \\ & \therefore \mathrm{x}+\mathrm{y}+1=\mathrm{c}(1-\mathrm{x}-\mathrm{y}-2 \mathrm{xy}) \\ & \end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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