Math, asked by gayathridevimj, 4 months ago

Solve dy / dx = 3x²y / 1- x³​

Answers

Answered by sanyamrewar
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.

Answered by mathdude500
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

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