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 $(x+y+1)=A(1-x–y-2xy)$

Answer 1

We have $(x+y+1)=A(1-x–y-2xy)$

Differentiate

$1+\frac{dy}{dx}=A(-1-\frac{dy}{dx}-2x\frac{dy}{dx}-2y)$

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

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

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

So, The general solution of the differential equation $\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$ is given by $(x+y+1)=A(1-x–y-2xy)$