dy/dx=x(2logx+1)/siny+ycosy
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Meenakshi Rajput 6 years, 9 months ago
- 1 answers
Related Questions
Posted by Arpit Sahu 1 month ago
- 0 answers
Posted by Abinav Singh 1 day, 22 hours ago
- 0 answers
Posted by Durgesh Kuntal Nishantkuntal 1 week, 6 days ago
- 0 answers
Posted by Pinky Kumari 2 weeks, 6 days ago
- 0 answers
Posted by Anupam Bind Anupam Bind 4 weeks, 2 days ago
- 0 answers
Posted by Suprith A 1 month ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sahdev Sharma 6 years, 9 months ago
{tex}{dy\over dx}={x(2log\ x+1)\over sin\ y+y.cos\ y}{/tex}
{tex} (sin \ y+y.cos\ y)dy = [x(2log\ x+1)]dx{/tex}
Integrating both sides, we get
{tex} \int sin\ y\ dy +\int y.cos\ y\ dy = \int 2x.log\ x \ dx+ \int xdx{/tex}
{tex}-cox \ y + y\ sin\ y + cos \ y = x^2log\ x -{x^2\over 2}+ {x^2\over 2}+C{/tex}
{tex}=> y sin\ y =x^2.log \ x+ C{/tex}
7Thank You