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

Q7.    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.

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best_answer

Given,
$\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\left(\frac{\mathrm{y}^2+\mathrm{y}+1}{\mathrm{x}^2+\mathrm{x}+1}\right)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^2+\mathrm{y}+1}=\frac{-\mathrm{dx}}{\mathrm{x}^2+\mathrm{x}+1}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^2+\mathrm{y}+1}+\frac{\mathrm{dx}}{\mathrm{x}^2+\mathrm{x}+1}=0$

Integrating both sides,

$\int \frac{d y}{y^2+y+1}+\int \frac{d x}{x^2+x+1}=C$
$\Rightarrow \int \frac{d y}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\int \frac{d y}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=C$
$\Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{\mathrm{y}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]+\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]=C$

$\Rightarrow \tan ^{-1}\left[\frac{2 y+1}{\sqrt{3}}\right]+\tan ^{-1}\left[\frac{2 x+1}{\sqrt{3}}\right]=\mathrm{C}$ $\Rightarrow \tan ^{-1}\left[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{2 y+1}{\sqrt{3}} \cdot \frac{2 x+1}{\sqrt{3}}}\right]=\frac{\sqrt{3}}{2} C$
$\Rightarrow \tan ^{-1}\left[\frac{\frac{2 x+2 y+2}{\sqrt{3}}}{1-\left(\frac{4 x y+2 x+2 y+1}{3}\right)}\right]=\frac{\sqrt{3}}{2} C$

$\Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}\right]=\frac{\sqrt{3}}{2} C$
$\Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}\right]=\frac{\sqrt{3}}{2} C$
$\Rightarrow \frac{\sqrt{3}(x+y+1)}{(1-x-y-2 x y)}=\tan \left(\frac{\sqrt{3}}{2} c\right)$
Let $\tan \left(\frac{\sqrt{3}}{2} c\right)=B$

$x+y+1=\frac{2 B}{\sqrt{3}}(1-x-y-2 x y)$
Let $A=\frac{2 B}{\sqrt{3}}$,
$x+y+1=A(1-x-y-2 x y)$

Hence proved.

Posted by

HARSH KANKARIA

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