Search any question & find its solution
Question:
Answered & Verified by Expert
The integrating factor of differential equation $\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0$ is
Options:
Solution:
1888 Upvotes
Verified Answer
The correct answer is:
$x$
(A)
We have $\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0$
$\therefore \frac{d y}{d x}=\frac{-\left(1+y+x^{2} y\right)}{x\left(1+x^{2}\right)}=\frac{-\left[1+y\left(1+x^{2}\right)\right]}{x\left(1+x^{2}\right)}$
$\therefore \frac{d y}{d x}=\frac{-1}{x\left(1+x^{2}\right)}-\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}$
$\therefore \frac{d y}{d x}+\left(\frac{1}{x}\right) y=\frac{-1}{x\left(1+x^{2}\right)}$
$\therefore$ I.F. $=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
We have $\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0$
$\therefore \frac{d y}{d x}=\frac{-\left(1+y+x^{2} y\right)}{x\left(1+x^{2}\right)}=\frac{-\left[1+y\left(1+x^{2}\right)\right]}{x\left(1+x^{2}\right)}$
$\therefore \frac{d y}{d x}=\frac{-1}{x\left(1+x^{2}\right)}-\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}$
$\therefore \frac{d y}{d x}+\left(\frac{1}{x}\right) y=\frac{-1}{x\left(1+x^{2}\right)}$
$\therefore$ I.F. $=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.