+91-8076753736 support@wiredfaculty.com
logo
logo

Sponsor Area

Vector Algebra

Question
CBSEENMA12033046
Wired Faculty App

For the following differential equation, find the general solution:
dy over dx space equals sin to the power of negative 1 end exponent straight x

Solution
The given differential equation is
                    dy over dx space equals space sin to the power of negative 1 end exponent straight x
Separating the variables and integrating,
                       integral space dy space equals space integral space sin to the power of negative 1 end exponent straight x space dx
therefore space space space space space space space space space space space space integral space 1 space dy space equals space integral space sin to the power of negative 1 end exponent straight x. space 1 space dx therefore space space space space space space space space space space straight y space equals space left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis. space straight x space minus space space integral fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction. straight x space dx or space space space space space space space space space space space space space space space space space space space straight y space equals space straight x space sin to the power of negative 1 end exponent straight x space plus space 1 half integral left parenthesis 1 minus straight x squared right parenthesis to the power of negative 1 half end exponent left parenthesis negative 2 space straight x right parenthesis space dx therefore space space space space space space space space space space space space space space space space space space space straight y space equals space straight x space sin to the power of negative 1 end exponent straight x space plus space 1 half fraction numerator left parenthesis 1 minus straight x squared right parenthesis to the power of begin display style 1 half end style end exponent over denominator begin display style 1 half end style end fraction plus straight c therefore space space space space space space space space space space space space space space space space space space space space straight y space equals space straight x space sin to the power of negative 1 end exponent straight x space plus space square root of 1 minus straight x squared end root plus straight c
which is the required solution.

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = ex

Determine order and degree (if defined) of differential equations:
y'' + (y')2 + 2 y = 0

Determine order and degree (if defined) of differential equations:
y″ + 2y′ + sin y = 0

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation