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

The given differential equation is dy dx + y^2+y+1 x^2+x+1 =0 or dy dx =- ( y^2+y+1 x^2+x+1 )=1 Integrating, we get dy y^2+y+1 =- dx x^2+x+1 dy y^2+y+1 + dx x^2+x+1 =0 dy (y+1 2 )^2+ (1-1 4 ) + dx ( x+1 2 )^2+1-1 4 =0 dy (y+1 2 )^2+ ( 3 2 )^2 + dx ( x+1 2 )^2+ ( 3 2 )^2 =0 2 3 ^ -1 ( y+1 2 3 2 )+2 3 ^ -1 ( x+1 2 3 2 )=0 ^ -1 ( 2y+1 3 )+ ^ -1 ( 2x+1 3 ) = 3C 2 =A_ 1 (say) Taking tangents on the two sides, we get 2y+1 3 + 2x+1 3 1- ( 2y+1 3 ) ( 2x+1 3 ) = _1 23(x+y+1) 3-(4xy+2x+2y+1) = _1 23(x+y+1) 2-(1-x-y-2xy) = _1 x+y+1=1 3 _1(1-x-y-2xy) x+y+1=A(1-x-y-2xy), where A=1 3 _1 is an arbitrary constant.

Show that the general solution of the differential equation dy dx…

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Question

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

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Answer

The given differential equation is
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}^2+\text{y}+1}{\text{x}^2+\text{x}+1}=0\ \ $ $\text{or}\ \ \frac{\text{dy}}{\text{dx}}=-\Big(\frac{\text{y}^2+\text{y}+1}{\text{x}^2+\text{x}+1}\Big)=1$
Integrating, we get
$\int\frac{\text{dy}}{\text{y}^2+\text{y}+1}=-\int\frac{\text{dx}}{\text{x}^2+\text{x}+1}$
$\int\frac{\text{dy}}{\text{y}^2+\text{y}+1}+\int\frac{\text{dx}}{\text{x}^2+\text{x}+1}=0$
$\Rightarrow\ \ \int\frac{\text{dy}}{\Big(\text{y}+\frac{1}{2}\Big)^2+\Big(1-\frac{1}{4}\Big)}+\int\frac{\text{dx}}{\Big({\text{x}+\frac{1}{2}\Big)^2+1-\frac{1}{4}}}=0$
$\Rightarrow\ \ \int\frac{\text{dy}}{\Big(\text{y}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}+\int\frac{\text{dx}}{\Big({\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}}\Big)^2}=0$
$\Rightarrow\ \ \frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{y}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)=0$
$\Rightarrow\ \ \tan^{-1}\Big(\frac{2\text{y}+1}{\sqrt{3}}\Big)+\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt{3}}\Big)$ $=\frac{\sqrt{3}\text{C}}{2}=\text{A}_{1}\ (\text{say})$
Taking tangents on the two sides, we get
$\frac{\frac{2\text{y}+1}{\sqrt{3}}+\frac{2\text{x}+1}{\sqrt{3}}}{1-\Big(\frac{2\text{y}+1}{\sqrt{3}}\Big)\Big(\frac{2\text{x}+1}{\sqrt{3}}\Big)}=\tan\text{A}_1$ $\Rightarrow\ \ \frac{2\sqrt{3}(\text{x}+\text{y}+1)}{3-(4\text{xy}+2\text{x}+2\text{y}+1)}=\tan\text{A}_1$
$\Rightarrow\ \ \frac{2\sqrt{3}(\text{x}+\text{y}+1)}{2-(1-\text{x}-\text{y}-2\text{xy})}=\tan\text{A}_1$
$\Rightarrow\ \ \text{x}+\text{y}+1=\frac{1}{\sqrt{3}}\tan\text{A}_1(1-\text{x}-\text{y}-2\text{xy})$
$\Rightarrow\ \ \text{x}+\text{y}+1=\text{A}(1-\text{x}-\text{y}-2\text{xy}),$
$\text{where A}=\frac{1}{\sqrt{3}}\tan\text{A}_1\ \text{is an arbitrary constant}.$

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