Question

xy^2(1+xy)dx+x^2y(1-2xy)dy=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential Equation is

$\rm :\longmapsto\: {xy}^{2}(1 + xy)dx \: + \: {yx}^{2} (1 - 2xy)dy = 0$

can be rewritten as

$\rm :\longmapsto\: y(1 + xy)dx \: + \: x (1 - 2xy)dy = 0 - - - (1)$

On comparing with,

$\bf :\longmapsto\:Mdx \: + \: Ndy = 0$

we get,

$\red{\rm :\longmapsto\:M = y(1 + xy) = y + {xy}^{2} } \\ \red{\rm :\longmapsto\:N = x(1 - 2xy) = x - {2yx}^{2} }$

Consider,

$\rm :\longmapsto\:\dfrac{\partial M}{\partial y} = 1 + 2xy$

and

$\rm :\longmapsto\:\dfrac{\partial N}{\partial x} = 1 - 4xy$

$\bf\implies \:\dfrac{\partial M}{\partial y} \ne \: \dfrac{\partial N}{\partial x}$

It means given equation is not exact Differential equation.

So, we have find Integrating Factor.

This Differential equation is of the form

$\rm :\longmapsto\:yf(xy)dx \: + \: xg(xy)dy = 0$

Its Integrating Factor, I.F. is

$\rm :\longmapsto\:I.F. = \dfrac{1}{Mx - Ny}$

$\rm \: \: = \: \: \dfrac{1}{x(y + {xy}^{2}) - y(x - {2yx}^{2})}$

$\rm \: \: = \: \: \dfrac{1}{xy + {x}^{2} {y}^{2} - yx + {2x}^{2} {y}^{2}}$

$\rm \: \: = \: \: \dfrac{1}{3 {x}^{2} {y}^{2} }$

$\bf\implies \:I.F. = \dfrac{1}{3 {x}^{2} {y}^{2} }$

Multiply equation (1) by I.F., we get

$\rm :\longmapsto\: \dfrac{y}{3 {x}^{2} {y}^{2} }(1 + xy)dx \: + \: \dfrac{x}{3 {x}^{2} {y}^{2} } (1 - 2xy)dy = 0$

$\rm :\longmapsto\:\bigg(\dfrac{1}{ {3yx}^{2} } + \dfrac{1}{3x} \bigg)dx + \bigg(\dfrac{1}{ {3xy}^{2} } - \dfrac{2}{3y} \bigg)dy = 0$

On comparing with,

$\bf :\longmapsto\:Mdx \: + \: Ndy = 0$

we get

$\red{\rm :\longmapsto\:M = \bigg(\dfrac{1}{ {3yx}^{2} } + \dfrac{1}{3x} \bigg)} \\ \red{\rm :\longmapsto\:N = \bigg(\dfrac{1}{ {3xy}^{2} } - \dfrac{2}{3y} \bigg)}$

Consider,

$\rm :\longmapsto\:\dfrac{\partial M}{\partial y} = \dfrac{ - 1}{3 {x}^{2} {y}^{2}}$

and

$\rm :\longmapsto\:\dfrac{\partial N}{\partial x} = - \dfrac{1}{3 {x}^{2} {y}^{2} }$

$\bf\implies \:\dfrac{\partial M}{\partial y} = \: \dfrac{\partial N}{\partial x}$

So equation is exact Differential equation,

Solution is given by

$\displaystyle \int_{taking \: y \: constant} \sf \: M \: dx + \int_{term \: not \: having \: x}Ndy = c$

$\displaystyle \int_{taking \: y \: constant} \sf \: \bigg(\dfrac{1}{ {3yx}^{2} } + \dfrac{1}{3x} \bigg) \: dx + \int_{term \: not \: having \: x}\bigg( - \dfrac{2}{3y} \bigg)dy = c$

$\rm :\longmapsto\: - \: \dfrac{1}{3xy} + \dfrac{1}{3} \: logx - \dfrac{2}{3} \: logy = c$