Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y = x sin x : xy' = y + x root of x^2 - y^2 ( x not equal to 0 and

6. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

       y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)

Answers (1)

best_answer

Given,

y = x\sin x

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(xsinx) = sinx + xcosx

Substituting the values of y' in LHS,

xy' = x(sinx + xcosx)

Substituting the values of y in RHS.

\\xsinx + x\sqrt{x^2 - x^2sin^2x} = xsinx + x^2\sqrt{1-sinx^2} = x(sinx+xcosx) = LHS

Therefore, the given function is a solution of the corresponding differential equation.

Posted by

HARSH KANKARIA

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 Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4
anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question