Question

Solve dy / dx = 3x²y / 1- x³​

Answer 1

Step-by-step explanation:

$\frac{dy}{dx} = \frac{3 {x}^{2}y }{1 - {x}^{3} } \\ \frac{dy}{y} = \frac{3 {x}^{2} }{1 - {x}^{3} } .dx$

Let 1-x³ =t

$- 3 {x}^{2} .dx = dt$

$\frac{dy}{y} = - \frac{dt}{t}$

intergarting both sides

$ln(y) = - ln(1 - {x}^{3} ) + c$

where c is constant of integration.

Answer 2

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

Given Differential equation is

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

Now, on separate the variables, we get

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

On integrating both sides, we get

$\rm :\longmapsto\: \displaystyle \int\dfrac{dy}{y} =\displaystyle \int \dfrac{3 {x}^{2} }{1 - {x}^{3} } \: dx$

can be rewritten as

$\rm :\longmapsto\: \displaystyle \int\dfrac{dy}{y} = - \displaystyle \int \dfrac{3 {x}^{2} }{{x}^{3} - 1} \: dx$

We know,

$\underbrace{ \boxed{ \bf \: \displaystyle \int \frac{1}{x}dx = logx + c}}$

and

$\underbrace{ \boxed{ \bf \: \displaystyle \int \frac{f'(x)}{f(x)} \: dx = logf(x) + c}}$

Also,

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

So, above integral reduced to

$\rm :\longmapsto\:logy = - log( {x}^{3} - 1) + logc$

$\rm :\longmapsto\:logy + log( {x}^{3} - 1) = logc$

We know,

$\underbrace{ \boxed{ \bf \: logx + logy = logxy}}$

So, using this

$\rm :\longmapsto\:logy({x}^{3} - 1) = logc$

$\bf\implies \:y( {x}^{3} - 1) = c$

Additional Information :-

Let's solve one question of same type!!

Question :- Solve the following Differential equation :-

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

Solution

Given Differential equation is

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

can be rewritten as

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

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle \int\dfrac{dy}{1 + {y}^{2} } =\displaystyle \int \dfrac{dx }{1 + {x}^{2} }$

We know,

$\underbrace{ \boxed{ \bf \: \displaystyle \int \frac{dx}{1 + {x}^{2} } = {tan}^{ - 1}x}}$

So, using this we get

$\rm :\longmapsto\: {tan}^{ - 1}y = {tan}^{ - 1}x + {tan}^{ - 1}c$

$\rm :\longmapsto\: {tan}^{ - 1}y - {tan}^{ - 1}x = {tan}^{ - 1}c$

$\rm :\longmapsto\: {tan}^{ - 1}\bigg(\dfrac{y - x}{1 + yx} \bigg) = {tan}^{ - 1}c$

$\bf :\longmapsto\: \dfrac{y - x}{1 + yx} = c$