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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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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} $

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Toolbox:
  • 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
 

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