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 satisfying the given condition$\large\frac{dy}{dx}$$+2y\tan x=\sin x;\;y=o\;when\;x=\large\frac{\pi}{3}$

$\begin{array}{1 1}(A)\;y = \cos x - 2\cos^2x \\ (B)\;y = \cos x + 2\cos^2x \\ (C)y = \sin x - 2\cos^2x \\(D)\;y = \cos x - 2\sin^2x \end{array} $

Can you answer this question?

1 Answer

0 votes
  • To solve the first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • (i) Write the given equation in the form of $\large\frac{dy}{dx}$$ + Py = Q$
  • (ii) Find the integrating factor (I.F) = $e^{\int Pdx}$.
  • (iii) Write the solution as y(I.F) = integration of $Q(I.F) dx + C$
Step 1:
Using the information in the tool box,
$\large\frac{dy}{dx}$$ +2y\tan x = \sin x$
Here $P = 2\tan x$ and $Q = \sin x$
To find the integral factor
$\int P dx = \int 2\tan xdx$
$\qquad\;\;=2 \log|\sec x|$ or $log|\sec^2x|$
Hence $I.F = e^{\large\log|\sec^2x|} = \sec^2x$
Step 2:
Hence the required solution is $y\sec^2x = \int\sin x(\sec^2x)dx + C$
$\int\sin x(\sec^2x)dx = \int\sec x.\tan x dx = \sec x$
Hence the require solution is $y\sec^2x = \sec x + C$
Step 3:
It is given $y = 0$ and $x = \large\frac{\pi}{3}$
substituting these values to evaluate C
$0.\sec^2(\large\frac{\pi}{3}) =$$ \sec(\large\frac{\pi}{3})$$ + C$
$\sec(\large\frac{\pi}{3})$$ = 2$
$c = -2$
Substituting the value of $C$ we get
$y\sec^2x = \sec x - 2$
dividing throughout by $\sec^2x$ we get
$y = \large\frac{1}{\sec x} - \frac{2}{\sec^2x}$
$y = \cos x - 2\cos^2x$
This is the required solution.
answered Jul 31, 2013 by sreemathi.v

Related questions

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