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Question:

$\frac{d y}{d x}+\sec x y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$

Solution:

The given differential equation is:

$\frac{d y}{d x}+p y=Q$ (where $p=\sec x$ and $Q=\tan x$ )

Now, I.F $=e^{\int p d x}=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int($ Q $\times$ I.F. $) d x+\mathrm{C}$

$\Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+\mathrm{C}$

$\Rightarrow y(\sec x+\tan x)=\int \sec x \tan x d x+\int \tan ^{2} x d x+\mathrm{C}$

$\Rightarrow y(\sec x+\tan x)=\sec x+\int\left(\sec ^{2} x-1\right) d x+\mathrm{C}$

$\Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+\mathrm{C}$

 

 

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