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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of the differential equation$\large\frac{dy}{dx}$$+(\sec x)y=\tan x\bigg(0\leq x<\large\frac{\pi}{2}\bigg)$

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

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  • 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:
The given equation is a first order linear differential equation.
Using the information in the tool box,
let us first find the I.F
$\int Pdx = \int \sec x = \log|\sec x + \tan x|$
Hence $I.F = e^{\Large\log|\sec x + \tan x| }= (\sec x + \tan x)$
Step 2:
The solution is $y.(\sec x+\tan x) = \int (\sec x + \tan x) dx + C$
$y(\sec x + \tan x) = \int\sec x\tan xdx + \int \tan^2xdx + C$
$y(\sec x+\tan x) = \sec x + \int (\sec^2x -1)dx + C$
$y(\sec x +\tan x) = \sec x + \tan x - x + C$
This is the required solution.
answered Aug 1, 2013 by sreemathi.v

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