Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The Integrating Factor of the differential equation ( 1 minus y to the power 2) dy by dx plus y x equals ay ( minus 1 less than y less than 1) is (A) (B) (C) (D)

Q19.    The Integrating Factor of the differential equation    (1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1) is

            (A)    \frac{1}{{y^2 -1}}

            (B)    \frac{1}{\sqrt{y^2 -1}}

            (C)    \frac{1}{{1 - y^2 }}

            (D)    \frac{1}{\sqrt{1 - y^2 }}    

Answers (1)

best_answer

Given equation is
(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)
we can rewrite it as
\frac{dx}{dy}+\frac{yx}{1-y^2}= \frac{ay}{1-y^2}
It is     \frac{dx}{dy}+px= Q   type of equation where p = \frac{y}{1-y^2}\ and \ Q = \frac{ay}{1-y^2}
Now,
I.F. = e^{\int pdy}= e^{\int \frac{y}{1-y^2}dy}= e^{\frac{\log |1 - y^2|}{-2}}= (1-y^2)^{\frac{-1}{2}}= \frac{1}{\sqrt{1-y^2}}
Therefore, the correct answer is (D)

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board exams twice a year from 2025, no plan for semesters: MoE
anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question