The solution of the differential equation ydx+(x+x^{2}y)dy=0  is

  • Option 1)

    \frac{1}{xy}+\log y=C\;

  • Option 2)

    \; -\frac{1}{xy}+\log y=C\;

  • Option 3)

    \;-\frac{1}{xy}=C\;

  • Option 4)

    \; \log y=Cx

 

Answers (1)
V Vakul

As we learnt in 

Bernoulli's Equation -

\frac{1}{y^{n-1}}= v

\frac{1}{y^{n}}\frac{dy}{dx}= \frac{1}{\left ( 1-n \right )}\frac{dv}{dx}

- wherein

\frac{1}{y^{n}}\frac{dy}{dx}+\frac{p}{y^{n-1}}=Q

 

 ydx+ (x+x^{2}y)dy=0

y\frac{dx}{dy}+ x+x^{2}y=0

\frac{dx}{dy}+ \frac{x}{y}+ x^{2} =0

Divided by x^{2}

\therefore \frac{1}{x^{2}}\frac{dx}{dy}+ \frac{1}{y.x}+1=0

Put \frac{1}{n}=t,\, \, -\frac{1}{x^{2}}\frac{dx}{dy}= \frac{dt}{dy}

-\frac{dt}{dy}+ \frac{t}{y}+1=0

p = -\frac{1}{y}, Q= 1

-\int \frac{1}{y} dy= -\log y= \log \frac{1}{y}

\Rightarrow I.F.= e^{\log \frac{1}{y}}= \frac{1}{y}

t.\frac{1}{y}= \int \frac{1}{y}dy = \log y

c+\frac{1}{xy}=log|y|

\therefore c= - \frac{1}{xy}+ \log \left | y \right |

 


Option 1)

\frac{1}{xy}+\log y=C\;

Incorrect option

Option 2)

\; -\frac{1}{xy}+\log y=C\;

Correct option

Option 3)

\;-\frac{1}{xy}=C\;

Incorrect option

Option 4)

\; \log y=Cx

Incorrect option

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