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    Q4.    \frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )

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Given equation is
\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )
This is  \frac{dy}{dx} + py = Q  type where p = \sec x and Q = \tan x
Now,
I.F. = e^{\int pdx}= e^{\int \sec xdx}= e^{\log |\sec x+ \tan x|}= \sec x+\tan x                     (\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sec x+\tan x) =\int ((\sec x+\tan x)\times \tan x)dx +C
y(\sec x+ \tan x) =\int (\sec x\tan x+\tan^2 x)dx +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
Therefore, the general solution is  y(\sec x+ \tan x) = \sec x +\tan x - x+C

Posted by

Gautam harsolia

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