The general solution of the differential equa | vedclass

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(C) Given the differential equation: $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ Separating the variables: $\frac{d y}{y^2+y+1} = -\frac{d x}{x^2+x+1}

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$x+y+1=c(1-x-y-2 x y)$

Vedclass · Answer

(C) Given the differential equation: $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ Separating the variables: $\frac{d y}{y^2+y+1} = -\frac{d x}{x^2+x+1}$ Completing the square in the denominators: $\int \frac{d y}{(y+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = -\int \frac{d x}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}$ Using the formula $\int \frac{du}{u^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + C$: $\frac{2}{\sqrt{3}} 	an^{-1}(\frac{2y+1}{\sqrt{3}}) = -\frac{2}{\sqrt{3}} 	an^{-1}(\frac{2x+1}{\sqrt{3}}) + C_1$ Dividing by $\frac{2}{\sqrt{3}}$ and rearranging: $	an^{-1}(\frac{2y+1}{\sqrt{3}}) + 	an^{-1}(\frac{2x+1}{\sqrt{3}}) = C_2$ Using the identity $	an^{-1} A + 	an^{-1} B = 	an^{-1}(\frac{A+B}{1-AB})$: $	an^{-1} \left[ \frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - (\frac{2y+1}{\sqrt{3}})(\frac{2x+1}{\sqrt{3}})} ight] = C_2$ Simplifying the expression inside: $\frac{\frac{2(x+y+1)}{\sqrt{3}}}{\frac{3 - (4xy+2x+2y+1)}{3}} = 	an C_2$ $\frac{2(x+y+1)}{\sqrt{3}} \cdot \frac{3}{2(1-x-y-2xy)} = 	an C_2$ $\frac{\sqrt{3}(x+y+1)}{1-x-y-2xy} = 	an C_2$ Thus,$x+y+1 = c(1-x-y-2xy)$,where $c = \frac{	an C_2}{\sqrt{3}}$.

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

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