Question

solve (x+2y^3)dy/DX=y​

Answer 1

Answer:

setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(4,0)(5,4)\qbezier(0,0)(0,0)(1,4)\qbezier(1,4)(1,4)(5,4)\qbezier(1,4)(1,4)(4,0)\put(-0.2,-0.4){\bf B}\put(0.8,4.2){\bf A}\put(4.2,-0.4){\bf C}\put(5.2,4.2){\sf D}\end{picture}

Answer 2

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

Given Differential equation is

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

can be rewritten as

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

can be rewritten as

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

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

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

This is a Linear Differential equation,

So,

On comparing with

$\red{\rm :\longmapsto\:\dfrac{dx}{dy} + px = q \: \: where \: p \: and \: q \: \in \: f(y)}$

We get,

$\rm :\longmapsto\:p \: = \: - \: \dfrac{1}{y}$

and

$\rm :\longmapsto\:q \: = \: {2y}^{2}$

So,

Integrating Factor is evaluated as

$\red{\rm :\longmapsto\:IF \: = {e} \: ^ {\displaystyle \int \rm \: pdy}}$

$\red{\rm :\longmapsto\:IF \: = {e} \: ^ {\displaystyle \int \rm \: - \: \frac{1}{y} dy}}$

$\red{\rm :\longmapsto\:IF \: = {e} \: ^ { - \: \displaystyle \int \rm \: \: \frac{1}{y} dy}}$

$\red{\rm :\longmapsto\:IF \: = {e} \: ^ { - \: logy}}$

$\red{\rm :\longmapsto\:IF \: = {e} \: ^ {\: log {y}^{ - 1} }}$

We know,

$\boxed{ \bf{ \: {e}^{logx} = x}}$

So, using this result, we get

$\red{\rm :\longmapsto\:IF = {y}^{ - 1}}$

$\bf\implies \:IF = \dfrac{1}{y}$

Now, Solution is given by

$\rm :\longmapsto\:x \times IF = \displaystyle \int \rm \: (q \times IF) \: dy$

$\rm :\longmapsto\:x \times \dfrac{1}{y} = \displaystyle \int \rm \: ( {2y}^{2} \times \frac{1}{y}) \: dy$

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

$\rm :\longmapsto \: \dfrac{x}{y} = 2 \times \dfrac{ {y}^{2} }{2} + c$

$\rm :\longmapsto \: \dfrac{x}{y} = {y}^{2} + c$

Additional Information :-

The linear differential equation of the form

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

Step :- 1 Integrating Factor

$\red{\rm :\longmapsto\:IF \: = {e} \: ^ {\displaystyle \int \rm \: pdx}}$

Step :- 2 Solution is given by

$\red{\rm :\longmapsto\:y \times IF = \displaystyle \int \rm \: (q \times IF) \: dx}$