Question

Show that the general solution of the differential equation $\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$ is given by $(x+y+1)=A(1-x-y-2xy)$, where $A$ is parameter.

Answer 1

$x+y+1=A(1-x-y-2xy)$

Differentiate both sides w.r.t. $x$

$\Rightarrow 1+y^{\prime }=A(-1-y^{\prime }-2y-2x{y}^{\prime })$

$\Rightarrow (1+A+2Ax)y^{\prime }=-1+A(-1-2y)$

$\Rightarrow y^{\prime }=-\frac{1+A(1+2y)}{1+A(1+2x)}$ .....$\left(1\right)$

Now, from the given equation, $A=\frac{x+y+1}{1-x-y-2xy}$.

Substitute this value of $x$ in equation $\left(1\right)$

$\Rightarrow y^{\prime }$ $=-\frac{1+\frac{(x+y+1)(1+2y)}{1-x-y-2xy}}{1+\frac{(x+y+1)(1+2x)}{1-x-y-2xy}}=-\frac{1-x-y-2xy+x+y+1+2xy+2y^2+2y}{1-x-y-2xy+x+y+1+2x^2+2xy+2x}=-\frac{2y^2+2y+2}{2x^2+2x+2}$

$\Rightarrow \frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$

Hence, $x+y+1=A(1-x-y-2xy)$ is the general solution of the given differential equation as it satisfies the equation.