Question

solve the differential equation (x^2y^2+xy)ydx+(x^2y^2-1)xdy=0​

Answer 1

Step-by-step explanation:

xy

2

(xy+1)dx+x(xy−1)(xy+1)dy=0

x=0x=0

y=0y=0

xy+1=0=>y=-{1 \over x}xy+1=0=>y=−

x

1

Or

y^2dx+(xy-1)dy=0y

2

dx+(xy−1)dy=0

Using an Integrating Factor

P(x,y)=y^2, Q(x,y)=xy-1P(x,y)=y

2

,Q(x,y)=xy−1

{\eth P \over \eth y}=2y, {\eth Q \over \eth x}=y, {\eth P \over \eth y}\not={\eth Q \over \eth x}

ðy

ðP

=2y,

ðx

ðQ

=y,

ðy

ðP



=

ðx

ðQ

{\eth P \over \eth y}-{\eth Q \over \eth x}=2y-y=y

ðy

ðP

−

ðx

ðQ

=2y−y=y

Integrating Factor

\mu=\mu(y)μ=μ(y)

{1 \over \mu}{d\mu \over dy}=-{1 \over P}({\eth P \over \eth y}-{\eth Q \over \eth x})

μ

1

dy

dμ

=−

P

1

(

ðy

ðP

−

ðx

ðQ

)

{1 \over \mu}{d\mu \over dy}=-{1 \over y^2}(2y-y)

μ

1

dy

dμ

=−

y

2

1

(2y−y)

{d\mu \over \mu}=-{dy \over y}

μ

dμ

=−

y

dy

\int {d\mu \over \mu}=\int (-{dy \over y})=>\ln|\mu|=\ln|{1 \over y}|+\ln C∫

μ

dμ

=∫(−

y

dy

)=>ln∣μ∣=ln∣

y

1

∣+lnC

We choose

\mu={1 \over y}μ=

y

1

Multiplying the original differential equation by \mu=1/yμ=1/y produces the exact equation:

ydx+(x-{1 \over y})dy=0ydx+(x−

y

1

)dy=0

Indeed, now we have

{\eth P \over \eth y}=1, {\eth Q \over \eth x}=1, {\eth P \over \eth y}={\eth Q \over \eth x}

ðy

ðP

=1,

ðx

ðQ

=1,

ðy

ðP

=

ðx

ðQ

Solve the resulting equation. The function

Answer 2

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

Given Differential equation is

$\rm :\longmapsto\:( {x}^{2} {y}^{2} + xy)ydx + ( {x}^{2} {y}^{2} - 1)xdy = 0$

On Comparing with Mdx + Ndy = 0, we get

$\rm :\longmapsto\:M \: \: = \: ( {x}^{2} {y}^{2} + xy)y = {x}^{2} {y}^{3} + {xy}^{2}$

and

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

Now,

Consider,

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

and

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

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

So, given equation is not exact Differential equation .

So, Consider again

$\rm :\longmapsto\:( {x}^{2} {y}^{2} + xy)ydx + ( {x}^{2} {y}^{2} - 1)xdy = 0$

So, Integrating factor is

$\rm \: = \: \:\dfrac{1}{Mx - Ny}$

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

$\rm \: = \: \:\dfrac{1}{xy(xy + 1)}$

Now, Multiply the given Differential equation by Integrating Factor we get .

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

$\rm :\longmapsto\:\: \:ydx + \dfrac{(xy + 1)(xy - 1)}{y(xy + 1)}dy = 0$

$\rm :\longmapsto\: \:ydx + \dfrac{(xy - 1)}{y}dy = 0$

$\rm :\longmapsto\: \:ydx +xdy - \dfrac{1}{y}dy = 0$

$\rm :\longmapsto\: \:d(xy) - \dfrac{1}{y}dy = 0$

On integrating both sides, we get

$\rm :\longmapsto\: \displaystyle\int\rm \:d(xy) - \displaystyle\int\rm \dfrac{1}{y}dy = 0$

$\rm :\longmapsto\:xy - logy = c$