For the given differential equation find the general solution.

$c o s^{2} d x \frac{d y}{d x} + y = t a n x (0 \leq x < \frac{p i}{2})$

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Solution

Given, $c o s^{2} d x \frac{d y}{d x} + y = t a n x (0 \leq x < \frac{p i}{2})$
On dividing by $c o s^{2} x$ on both sides, we get
$\frac{d y}{d x} + y s e c^{2} x = t a n x s e c^{2} x$
On comparing with the form $\frac{d y}{d x} + P y = Q$ we get P=secx and $Q = t a n x$
$∴ I F = e^{\int P d x} \Rightarrow e^{\int s e c^{2} x d x} = I F = e^{t a n x}$ ...(i)
The general solution of the given differential equation is given by
$y . I F = \int Q \times I F d x + C \Rightarrow y e^{t a n x} = \int e^{t a n x} . t a n x s e c^{2} x d x P u t t a n x = t \Rightarrow s e c^{2} x = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{s e c^{2} x}$
$∴ y e^{t a n x} = \int e^{t} . t s e c^{2} x \frac{d t}{s e c^{2} x} \Rightarrow y e^{t a n x} = \int e^{t} . t d x \Rightarrow y e^{t a n x} = e^{t} . t - \int [\frac{d}{d t} (t) . \int e^{t} d t] d t \Rightarrow y e^{t a n x} = e^{t} . t - \int e^{t} d t \Rightarrow y e^{t a n x} = e^{t} t - e^{t} + C \Rightarrow y e^{t a n x} = (t a n x - 1) e^{t a n x} + C \Rightarrow y = t a n x - 1 + C e^{- t a n x}$

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