Question

$y(1 + xy + {x}^{2} {y}^{2} )dx \: + x(1 - xy + {x}^{2} {y}^{2} )dy \: = 0$
​

Answer 1

Solution :

$\sf{y(1 + xy + x^{2}y^{2})dx + x(1 - xy + x^{2}y^{2})dy = 0}$

$\sf{let\:t = xy, then\:y = \dfrac{t}{x}\:and\: dy = \dfrac{xdt - tdx}{x^{2}}}$

$\boxed{\begin{minipage}{8 cm} \sf{\underline{To\:find\:dy = \dfrac{xdt - tdx}{x^{2}}}} : \\ \\ \\ \sf{:\implies \dfrac{dy}{dx} = \dfrac{d}{dx}\bigg(\dfrac{t}{x}\bigg)} \\ \\ \textsf{By using the quotient rule of differentiation, we get :} \\ \\ \sf{:\implies \dfrac{dy}{dx} = \dfrac{x\dfrac{dt}{dx} - t\dfrac{dx}{dx}}{x^{2}}} \\ \\ \sf{:\implies dy = \dfrac{xdt - tdx}{dx \times x^{2}} \times dx} \\ \\ \sf{:\implies dy = \dfrac{xdt - tdx}{x^{2}}} \\ \\ \sf{Hence,\:dy = \dfrac{xdt - tdx}{x^{2}}} \end{minipage}}$

By substituting the value of xy, y and dy in the equation, we get :

$:\implies \sf{\bigg[\bigg(\dfrac{t}{x}\bigg)\bigg(1 + t + t^{2}\bigg)dx\bigg] + \bigg[x\bigg(1 - t + t^{2}\bigg)\bigg(\dfrac{xdt - tdx}{x^{2}}\bigg)\bigg] = 0}$

$:\implies \sf{\bigg[\bigg(\dfrac{t}{x}\bigg)\bigg(1 + t + t^{2}\bigg)dx\bigg] + \bigg[x\bigg(1 - t + t^{2}\bigg)\bigg(\dfrac{dt}{x} - \dfrac{tdx}{x^{2}}\bigg)\bigg] = 0}$

$:\implies \sf{\bigg[\bigg(\dfrac{t}{x}\bigg)\bigg(1 + t + t^{2}\bigg)dx - \bigg(\dfrac{t}{x}\bigg)\bigg(1 - t + t^{2}\bigg)dx\bigg] = \bigg(t - t^{2} - 1\bigg)dt}$

$:\implies \sf{\bigg[\bigg(\dfrac{t}{x}\bigg)\bigg\{\bigg(1 + t + t^{2}\bigg) - \bigg(1 - t + t^{2}\bigg)\bigg\}\bigg]dx = \bigg(t - t^{2} - 1\bigg)dt}$

$:\implies \sf{\bigg[\bigg(\dfrac{t}{x}\bigg)\bigg(1 + t + t^{2} - 1 + t - t^{2}\bigg)\bigg]dx = \bigg(t - t^{2} - 1\bigg)dt}$

$:\implies \sf{\bigg[\bigg(\dfrac{t}{x}\bigg)\bigg(2t\bigg)\bigg]dx = \bigg(t - t^{2} - 1\bigg)dt}$

$:\implies \sf{\bigg(\dfrac{2t^{2}}{x}\bigg)dx = \bigg(t - t^{2} - 1\bigg)dt}$

$:\implies \sf{\bigg(\dfrac{1}{x}\bigg)dx = \bigg(\dfrac{t - t^{2} - 1}{2t^{2}}\bigg)dt}$

By integrating on both the sides, we get :

$:\implies \displaystyle\int \sf{\bigg(\dfrac{1}{x}\bigg)dx} = \displaystyle\int \sf{\bigg(\dfrac{t - t^{2} - 1}{2t^{2}}\bigg)dt}$

$:\implies \displaystyle\int \sf{\bigg(\dfrac{1}{x}\bigg)dx} = \displaystyle\int \sf{\bigg(\dfrac{t}{2t^{2}}\bigg)dt} - \displaystyle\int \sf{\bigg(\dfrac{t^{2}}{2t^{2}}\bigg)dt} - \displaystyle\int \sf{\bigg(\dfrac{1}{2t^{2}}\bigg)dt}$

$:\implies \sf{\bigg(In|x|\bigg) = \dfrac{1}{2}\bigg(In|t|\bigg) - \dfrac{1}{2}\bigg(t\bigg) + \bigg(\dfrac{1}{2t}\bigg) + C}$

Now, by substituting the value of t in the equation, we get :

$:\implies \sf{\bigg(In|x|\bigg) = \dfrac{1}{2}\bigg(In|xy|\bigg) - \dfrac{1}{2}\bigg(xy\bigg) + \bigg(\dfrac{1}{2xy}\bigg) + C}$

$\boxed{\therefore \sf{\bigg(In|x|\bigg) = \dfrac{1}{2}\bigg(In|xy|\bigg) - \dfrac{1}{2}\bigg(xy\bigg) + \bigg(\dfrac{1}{2xy}\bigg) + C}}$

Answer 2

✯ Answer ✯

$dy/dx =y (1+xy+x^2y^2/ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x(1-xy+x^2y^2)$

$Let \: U=xy$

$then \: y = u/x => dy= (xdy - udx/x^2)$

$(x \: \: \: du - udx) /x^2 dx = \\ \: \: \: \: \: \: \: \: \: y/x (1+u+u^2) /(1-u+u^2)$

$xdu - udx/dx = u (1+u+u^2) / \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1-u+u^2$

$we \: get,$

$x \: \: \: \: \: du/dx = u(1+u+u^2) + \\ \: \: \: \: \: \: \: \: \: \: \: \: u(1-u+u^2) /1-u + u^2$

$x \: \: \: \: du/dx = 2(u+u^3) /(1-u+u^2)$

$du ( 1 - u+u^2/2(u+u^3) )= dx/x$

$Integrate \: \: both \: \: side$

$ln |x| = -1/2 \: u+1/2 \: ln |u| + 1/2 u^2 + c$

$u= xy$

$ln |x| = - xy/ 2 + 1/2 \: ln | xy| + 1/ xy + c$

Step-by-step explanation:

$@BrainlyGENIUS$