Question

solve differential equation dy/dx+y^2+y+1/x^2+x+1=0​

Answer 1

EXPLANATION.

$\sf \implies \dfrac{dy}{dx} \ + \dfrac{y^{2} + y + 1 }{x^{2} + x + 1} = 0.$

As we know that,

$\sf \implies \dfrac{dy}{dx} \ = -\bigg(\dfrac{y^{2} + y + 1}{x^{2} + x + 1} \bigg)$

$\sf \implies \dfrac{dy}{y^{2} + y + 1} \ = -\bigg(\dfrac{dx}{x^{2} + x + 1} \bigg)$

Integrate both sides, we get.

$\sf \implies \displaystyle \int \dfrac{dy}{y^{2} + y + 1} \ = \int \dfrac{-dx}{x^{2} + x + 1}$

Integrate individually, we get.

$\sf \implies \displaystyle \int \dfrac{dy}{y^{2} + y + 1}$

$\sf \implies \displaystyle \int \dfrac{dy}{\bigg(y + \dfrac{1}{2} \bigg)^{2} + \dfrac{3}{4} }$

$\sf \implies \displaystyle \int \dfrac{dy}{\bigg(y + \dfrac{1}{2} \bigg)^{2} + \bigg(\dfrac{\sqrt{3} }{2} \bigg)^{2} }$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} + a^{2} } \ = \dfrac{1}{a} tan^{-1} \bigg(\dfrac{x}{a} \bigg) + C$

Using this formula in equation, we get.

$\sf \implies \displaystyle \dfrac{1}{\bigg(\dfrac{\sqrt{3} }{2}\bigg) } tan^{-1} \bigg(\dfrac{y + \dfrac{1}{2} }{\dfrac{\sqrt{3} }{2} } \bigg) + c.$

$\sf \implies \displaystyle \dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2y + 1}{\sqrt{3} } \bigg) + C$

$\sf \implies \displaystyle \int \dfrac{dy}{y^{2} + y + 1} = \dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2y + 1}{\sqrt{3} } \bigg) + C_{1}.$

Same we integrate other equation, we get.

$\sf \implies \displaystyle \int \dfrac{-dx}{x^{2} + x + 1}$

$\sf \implies \displaystyle \int \dfrac{-dx}{\bigg(x + \dfrac{1}{2} \bigg)^{2} + \dfrac{3}{4} }$

$\sf \implies \displaystyle \int \dfrac{-dx}{\bigg( x + \dfrac{1}{2} \bigg)^{2} + \bigg(\dfrac{\sqrt{3} }{2}\bigg)^{2} }$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} + a^{2} } \ = \dfrac{1}{a} tan^{-1} \bigg(\dfrac{x}{a} \bigg) + C$

Using this formula in equation, we get.

$\sf \implies \displaystyle- \dfrac{1}{\bigg(\dfrac{\sqrt{3} }{2} \bigg)} tan^{-1} \bigg(\dfrac{x + \dfrac{1}{2} }{\dfrac{\sqrt{3} }{2} } \bigg) + C$

$\sf \implies \displaystyle -\dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2x + 1}{\sqrt{3} } \bigg) + C$

$\sf \implies \displaystyle \int \dfrac{-dx}{x^{2} + x + 1} \ =- \dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2x + 1}{\sqrt{3} }\bigg) + C_{2}.$

$\sf \implies \displaystyle \dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2y + 1}{\sqrt{3} } \bigg) + C_{1} \ = - \displaystyle \dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2y + 1}{\sqrt{3} } \bigg) + C_{2}$

$\sf \implies \displaystyle tan^{-1} \bigg[\dfrac{2y + 1}{\sqrt{3} } \bigg] \ + tan^{-1} \bigg[ \dfrac{2x + 1}{\sqrt{3} } \bigg] = \dfrac{\sqrt{3} C}{2}.$

As we know that,

Formula of :

$\sf \implies \displaystyle tan^{-1} (x) \ + tan^{-1} (y) \ = tan^{-1} \bigg[\dfrac{x + y}{1 - x.y} \bigg]$

Using this formula in equation, we get.

$\sf \implies \displaystyle tan^{-1} \bigg[ \dfrac{\dfrac{2y + 1}{\sqrt{3} } \ + \dfrac{2x + 1}{\sqrt{3} } }{1 - \bigg(\dfrac{2y + 1}{\sqrt{3}}\bigg)\bigg(\dfrac{2x + 1}{\sqrt{3} } \bigg) } \bigg] \ = \dfrac{\sqrt{3} C}{2}$

$\sf \implies \displaystyle tan^{-1} \bigg[ \dfrac{\dfrac{2x + 2y + 1}{\sqrt{3} } }{1 - \bigg(\dfrac{4xy + 2x + 2y + 1}{3} \bigg)} \bigg] = \dfrac{\sqrt{3}C }{2}$

$\sf \implies \displaystyle tan^{-1} \bigg[ \dfrac{2\sqrt{3}(x + y + 1) }{3 - 4xy - 2x - 2y - 1} \bigg] = \dfrac{\sqrt{3} C}{2}$

$\sf \implies \displaystyle tan^{-1} \bigg[ \dfrac{\sqrt{3} (x + y + 1)}{2(1 - x - y - 2xy)} \bigg] = \dfrac{\sqrt{3} C}{2}$

$\sf \implies \displaystyle \bigg[ \dfrac{\sqrt{3}(x + y + 1) }{2(1 - x - y - 2xy)} \bigg] = tan\bigg[ \dfrac{\sqrt{3} C}{2} \bigg]$