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  • Find the particular solution of the differential equation , where <img alt="x\in \begin{bmatrix}

Find the particular solution of the differential equation \frac{dy}{dx}+y \sec x=\tan x, where x\in \begin{bmatrix} 0 &\frac{\pi}{2} \end{bmatrix},

given that y = 1, when x=\frac{\pi}{4}

 

 
 
 
 
 

Answers (1)

\frac{dy}{dx}+y \sec x=\tan x

P= \sec x

I.F. = e^{ \int \sec x dx } = e^{ \log( \sec x + \tan x)} = \sec x + \tan x

y ( \sec x + \tan x) = \int (\sec x + \tan x)\tan x dx

y ( \sec x + \tan x) = \int \sec x\tan x dx + \int \tan^2 x dx

y ( \sec x + \tan x) = \sec x + \int( 1+ \sec^2 x) dx

y ( \sec x + \tan x) = \sec x + x + \tan x +c

y = 1 at  x=\frac{\pi}{4}

1( \sec \frac{\pi}{4} + \tan \frac{\pi}{4}) = \sec \frac{\pi}{4} + \frac{\pi}{4} + \tan \frac{\pi}{4} +c

c = - \frac{\pi}{4}

Particular solution:

y ( \sec x + \tan x) = \sec x + x + \tan x -\frac{\pi}{4}

Posted by

Safeer PP

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