The integrating factor of the differential equation (xy-1)\frac{dy}{dx}+y^{2}=0 is

  • Option 1)

    y

  • Option 2)

    \frac{1}{y}

  • Option 3)

    \frac{1}{xy}

  • Option 4)

    xy

 

Answers (1)

 

Bernoulli's Equation -

\frac{dy}{dx}+py =Qy^{n}

- wherein

P,Q are the function of x alone.

 

 (xy-1)\frac{dy}{dx}+y^{2}=0

\therefore- \frac{1}{y^{2}}\frac{dy}{dx}=\frac{1}{xy-1}

put\: \frac{1}{y}=\frac{1}{t}

\therefore- \frac{1}{y^{2}}\frac{dy}{dx}=\frac{dt}{dx}

\frac{dt}{dx}=\frac{1}{\frac{x}{t}-1}

\left [ \because y=\frac{1}{t} \right ]

\frac{dx}{dt}=\frac{x}{t}-1

\frac{dx}{dt}-\frac{x}{t}=-1

p=-\frac{1}{t}

\therefore \int Pdt=-\int \frac{1}{t}dt

=-logt

=log\frac{1}{t}

I.f=e^{log\frac{1}{t}}

=>\frac{1}{t}=y

 


Option 1)

y

Option is correct

Option 2)

\frac{1}{y}

Option is incorrect

Option 3)

\frac{1}{xy}

Option is incorrect

Option 4)

xy

Option is incorrect

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions