Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter 21 Differential Equation exercise Fill in the blank question 17 maths textbook solution

Please solve RD Sharma class 12 chapter 21 Differential Equation exercise Fill in the blank question 17 maths textbook solution

Answers (1)

Answer:

 \frac{e^{x}}{x}

Hint:

 Use the form

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q,

to find the integrating factor.

Given:

 \frac{\mathrm{d} y}{\mathrm{d} x}+y=\frac{1+y}{x}

Solution:

\frac{\mathrm{d} y}{\mathrm{d} x}+y-\frac{y}{x}=\frac{1}{x}

\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+y\left ( 1-\frac{1}{x} \right )=\frac{1}{x}

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q,

Where

P=\left ( 1-\frac{1}{x} \right )

I.F=e^{\int P\, dx}=e^{\int(1-\frac{1}{x}) dx}=e^{(x-in\, x)}

I.F=e^{x}.e^{-in\, x}=e^{x}.e^{in_{e}\, (x^{-1})}=e^{x}.\frac{1}{x}=\frac{e^{x}}{x}

So, the answer is

\frac{e^{x}}{x}

Posted by

Gurleen Kaur

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

‘You ruined my career’: Students hit back at CBSE over 'abrupt' removal of additional subjects
anam-khan

‘Careers are at stake’: CBSE removes additional subject option for private students without prior notice
anam-khan

CBSE eases APAAR ID mandate for board exams 2026 over lack of parental consent, technical issues

Latest Question