Question

solve differential equation dy/dx+x*y=x*y^3​

Answer 1

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

Given Differential equation is

$\rm :\longmapsto\:\dfrac{dy}{dx} + xy = {xy}^{3}$

$\red{\rm :\longmapsto\:Divide \: both \: sides \: by \: {y}^{3}}$

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

$\red{\rm :\longmapsto\:Let \: \dfrac{1}{ {y}^{2} } = t}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto \: \dfrac{ - 2}{ {y}^{3} } \dfrac{dy}{dx}= \dfrac{dt}{dx}}$

$\red{\rm :\longmapsto \: \dfrac{1}{ {y}^{3} } \dfrac{dy}{dx}= \dfrac{ - 1}{2} \dfrac{dt}{dx}}$

So, equation (1) can be rewritten as

$\rm :\longmapsto\:\dfrac{ - 1}{2}\dfrac{dt}{dx} +tx = x$

$\rm :\longmapsto\:\dfrac{dt}{dx} - 2tx = - 2x$

Its a linear differential equation,

On comparing with

$\red{\rm :\longmapsto\:\dfrac{dt}{dx} + pt = q \: \: where \: p \: and \: q \: \in \: f(x)}$

we get

$\rm :\longmapsto\:p = - 2x$

$\rm :\longmapsto\:q = - 2x$

Now,

Integrating factor, is

$\rm :\longmapsto\:IF \: = \: {e} \: ^{\displaystyle\int\tt \: p \: dx}$

$\rm :\longmapsto\:IF \: = \: {e} \: ^{\displaystyle\int\tt \: - 2x \: dx}$

$\rm :\longmapsto\:IF \: = \: {e} \: ^{ - {x}^{2} }$

Solution of differential equation is

$\rm :\longmapsto\:t \times IF = \displaystyle\int\tt \: (q \times IF) \: dx$

$\rm :\longmapsto\:t \times {e}^{ - {x}^{2} } = \displaystyle\int\tt \: ( - 2x \times {e}^{ - {x}^{2} } ) \: dx$

$\rm :\longmapsto\:t \times {e}^{ - {x}^{2} } = {e}^{ - {x}^{2} } + c$

$\: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \displaystyle\int\tt {e}^{f(x)} \: f'(x) \: dx = {e}^{f(x)} + c \bigg \}}$

$\rm :\implies\:t = 1 + c {e}^{{x}^{2} }$

On substituting the value of t, we get

$\rm :\implies\:\dfrac{1}{ {y}^{2} } = 1 + c {e}^{{x}^{2} }$

$\rm :\longmapsto\: {y}^{2} + {cy}^{2} {e}^{ {x}^{2} } = 1$