Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Find a particular solution of the differential equation $\frac{dy}{dx} + y\cot x=4x\:cosec\:x$, $(x\neq0)$, given that $y = 0$ when $x = \large\frac{\pi}{2}$

Can you answer this question?

1 Answer

0 votes
  • To solve the first order linear differential equation:
  • (i) First write the equation of the form $\large\frac{dy}{dx}$$ + py = Q$
  • (ii) Find the integral factor(I.F) =$ e^{\int pdx}$
  • (iii) Write the solution as y.(I.F) - integration of Q.(I.F) + C
Step 1:
Given $\large\frac{dy}{dx }$$+y\cot x = 4xcosec x $
This is a first order linear differential equation.
Here $p = \cot x$ and $Q = 4xcosec\;x$
$\int pdx=\int \cot x=\log\mid\sin x\mid$
Hence $I.F=e^{\large\log\mid\sin x\mid}=\sin x$
Step 2:
Hence the solution is $y\sin x = \int4x cosec\;x.\sin x dx + C$
$cosec x=\large\frac{1}{\sin x}$
$y\sin x=\int 4xdx+C$
$y\sin x=\large\frac{4x^2}{2}$$+C$
$y\sin x=2x^2+C$-----(1)
Step 3:
To evaluate the value of C, let us substitute $x = \large\frac{\pi}{2}$ and $y = 0 $
Substituting the value of C in equ(1) we get,
$y\sin x=2x^2-\large\frac{\pi^2}{2}$
This is the required solution.
answered Jul 29, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App