Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find a particular solution of the differential equation dy/dx + y cot x = 4x cosec x (x not equal to 0), given that y = 0 when x = pi/2

Q13.    Find a particular solution of the differential equation \frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0), given that y = 0 \ \textup{when}\ x = \frac{\pi}{2}.

Answers (1)

best_answer

Given equation is
\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)
This is  \frac{dy}{dx} + py = Q  type where p =\cot x and Q = 4xcosec xQ = 4x \ cosec x
Now,
I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x                     
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C
y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C\\ \\ y(\sin x) = \int 4x + C\\ y\sin x= 2x^2+C
Now, by using boundary conditions we will find the value of C
It is given that  y = 0 when x= \frac{\pi}{2}
at   x= \frac{\pi}{2}
0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C\\ \\ C = - \frac{\pi^2}{2}
Now, put the value of C
y\sin x= 2x^2-\frac{\pi^2}{2}
Therefore, the particular solution is  y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0) 

Posted by

Gautam harsolia

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question