Question

(1+xy)ydx+(1-xy)xdy=0​

Answer 1

Given Question :-

• Solve the differential equation

$\sf \: (1 + xy)ydx + (1 - xy)xdy = 0$

$\large\underline{ \bf{Solution :- }}$

• Given Differential equation is

$\rm :\longmapsto\:(1 + xy)ydx + (1 - xy)xdy = 0$

• On comparing with Mdx + Ndy = 0,

we get

• M = (1 + xy)y

and

• N = (1 - xy)x

Now,

• Consider

$\rm :\longmapsto\:\dfrac{ \partial \:M }{\partial \:y} = \dfrac{\partial \:}{\partial \:y} (y + {xy}^{2} )$

$\rm :\implies\:\dfrac{\partial \:M}{\partial \:y} = 1 + 2xy - - (1)$

Now,

• Consider,

$\rm :\longmapsto\:\dfrac{\partial \:N}{\partial \:x} = \dfrac{\partial \:}{\partial \:x} (x - {yx}^{2} )$

$\rm :\implies\:\dfrac{\partial \:N}{\partial \:x} = 1 - 2xy - - (2)$

• So, from equation (1) and equation (2), we concluded

$\rm :\longmapsto\:\dfrac{\partial \:M}{\partial \:y} \ne \: \dfrac{\partial \:N}{\partial \:x}$

So,

• Given Differential equation is not Exact Differential Equation.

• To solve the given Differential equation, we use inspection method

The given Differential equation can be rewritten as

$\rm :\longmapsto\:ydx + {xy}^{2}dx + xdy - {yx}^{2}dy = 0$

$\rm :\longmapsto\:(ydx + xdy) + xy(ydx - xdy) = 0$

$\rm :\longmapsto\:d(xy) + xy(ydx - xdy) = 0$

$\: \: \boxed{ \bf \: Divide \: whole \: equation \: by \: {x}^{2} {y}^{2} }$

$\rm :\longmapsto\:\dfrac{d(xy)}{ {(xy)}^{2} } + \dfrac{ydx - xdy}{xy} = 0$

$\rm :\longmapsto\:\dfrac{d(xy)}{ {(xy)}^{2} } + \dfrac{ydx}{xy} - \dfrac{xdy}{xy} = 0$

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

Now, on integrating both

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

$\rm :\longmapsto\: - \: \dfrac{1}{xy} + log(x) - log(y) = c$

$\boxed{ \bf \: \because \: \int\dfrac{dx}{ {x}^{2} } = - \dfrac{1}{x} + c \: \: \: and \: \: \: \int\dfrac{dx}{x} = log(x) + c}$

$\bf\implies \: log \bigg(\dfrac{x}{y} \bigg) = c + \dfrac{1}{xy}$

Answer 2

We are given to solve the differential equation,

$\longrightarrow(1+xy)y\ dx+(1-xy)x\ dy=0\quad\quad\dots(1)$

Let,

• $x=\dfrac{1}{u}$

• $dx=-\dfrac{1}{u^2}\ du$

and,

• $y=\dfrac{1}{v}$

• $dy=-\dfrac{1}{v^2}\ dv$

Then (1) becomes,

$\longrightarrow-\left(1+\dfrac{1}{uv}\right)\dfrac{1}{v}\cdot\dfrac{1}{u^2}\ du-\left(1-\dfrac{1}{uv}\right)\dfrac{1}{u}\cdot\dfrac{1}{v^2}\ dv=0$

$\longrightarrow\left(1+\dfrac{1}{uv}\right)\dfrac{1}{u}\ du+\left(1-\dfrac{1}{uv}\right)\dfrac{1}{v}\ dv=0$

$\longrightarrow\left(\dfrac{uv+1}{uv}\right)\dfrac{1}{u}\ du+\left(\dfrac{uv-1}{uv}\right)\dfrac{1}{v}\ dv=0$

$\longrightarrow\left(uv+1\right)\dfrac{1}{u}\ du+\left(uv-1\right)\dfrac{1}{v}\ dv=0$

$\longrightarrow v\ du+\dfrac{1}{u}\ du+u\ dv-\dfrac{1}{v}\ dv=0$

$\longrightarrow v\ du+u\ dv+\dfrac{1}{u}\ du-\dfrac{1}{v}\ dv=0$

$\longrightarrow d(uv)+\dfrac{1}{u}\ du-\dfrac{1}{v}\ dv=0$

Now integrating,

$\displaystyle\longrightarrow\int d(uv)+\int\dfrac{1}{u}\ du-\int \dfrac{1}{v}\ dv=C$

$\displaystyle\longrightarrow uv+\log|u|-\log|v|=C$

$\displaystyle\longrightarrow uv+\log\left|\dfrac{u}{v}\right|=C$

Undoing substitutions $x=\dfrac{1}{u}$ and $y=\dfrac{1}{v},$

$\displaystyle\longrightarrow\underline{\underline{\dfrac{1}{xy}+\log\left|\dfrac{y}{x}\right|=C}}$