No products in the cart.

Find the particular solution of the …

CBSE, JEE, NEET, CUET

CBSE, JEE, NEET, CUET

Question Bank, Mock Tests, Exam Papers

NCERT Solutions, Sample Papers, Notes, Videos

Find the particular solution of the diffrential equation xdx-ye^y √1+x² dy=0, given that y=1 when x=0
  • 1 answers

Mp Raju 5 years, 6 months ago

given d.e

xdx-yey{tex}\sqrt{1+x^2} {/tex}dy=0

both sides divided by {tex}\sqrt{1+x^2}{/tex}

{tex}\implies {/tex}{tex}\frac{x}{\sqrt{1+x^2}} {/tex}dx-y ey dy=0

{tex}\implies {/tex}{tex}\int{/tex} {tex}\frac{x}{\sqrt{1+x^2}}dx-\int y e^y dy =0 {/tex}

{tex}\implies {/tex}{tex}\sqrt{1+x^2}{/tex} - yey+ey=c

at (0,1)

{tex}\implies{/tex}1-e+e=c

{tex}\implies{/tex}c=1

particular solution is {tex}\sqrt{1+x^2}{/tex}{tex}- ye^y +e^y=1{/tex}

http://mycbseguide.com/examin8/

Related Questions

Y=sin√ax^2+√bx+√c
  • 0 answers
Three friends Ravi Raju
  • 0 answers

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App